Question 15 Marks
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer Let A be any point on the circle of reference of the figure From A, draw BN perpendicular on x-axis.
If $\angle\text{POA}=\theta,$ then
$\angle\text{OAM}=\theta=\omega\text{t}$
$\therefore$ In triangle OAM,
$\frac{\text{OM}}{\text{OA}}=\sin\theta$
$\therefore\ \frac{-\text{x}}{3}=\omega\text{t}=\frac{\sin(2\pi)}{\text{T}}\text{t}$
$\therefore\ \text{x}=-3\frac{\sin(2\pi)}{2}\text{t}$ or $\text{x}=-3\sin\pi\text{t}$ which is the equation of SHM.
View full question & answer→Question 25 Marks
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer Let B be any point on the circle of reference of the figure. From B draw BN perpendicular on x-axis. Then $\triangle\text{BON}=\theta=\omega\text{t}$ $\therefore$ In $\triangle\text{ONB},\ \cos\theta=\frac{\text{ON}}{\text{OB}}$ Or $\text{ON}=\text{OB}\cos\theta$$\therefore\ -\text{x}=2\cos\omega\text{t}$
$\Rightarrow\ \text{x}=-2\frac{\cos(2\pi)}{\text{T}}\text{t}=-2\cos\frac{2\pi}{4}\text{t}$
$\therefore\ \text{x}=-2\cos\frac{\pi}{4}\text{t}$ which is equation of SHM
View full question & answer→Question 35 Marks
The acceleration due to gravity on the surface of moon is $1.7m s^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is $3.5 s$? (g on the surface of earth is $9.8m s^{-2}$)
AnswerAcceleration due to gravity on the surface of moon, $g' = 1.7m s^{–2}$ Acceleration due to gravity on the surface of earth, $g = 9.8m s^{–2}$ Time period of a simple pendulum on earth, T = 3.5s $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ Where l is the length of the pendulum $\therefore\ \text{l}=\frac{\text{T}^2}{(2\pi)^2}\times\text{g}$
$=\frac{(3.5)^2}{4\times(3.14)^2}\times9.8\text{m}$ The length of the pendulum remains constant. On moon's surface, time period, $\text{T}'=2\pi\sqrt{\frac{\text{l}}{\text{g}'}}$
$=2\pi\sqrt{\frac{\frac{(3.5)^2}{(4\times3.14)^2}\times9.8}{1.7}}=8.4\text{s}$Hence, the time period of the simple pendulum on the surface of moon is 8.4s.
View full question & answer→Question 45 Marks
A mass attached to a spring is free to oscillate, with angular velocity $\omega,$ in a horizontal plane without friction or damping. It is pulled to a distance $x_0$ and pushed towards the centre with a velocity $\upsilon_0$ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega,x_0$ and $υ_0$. [Hint: Start with the equation $\text{x}=\text{a}\cos(\omega\text{t}+\theta)$ and note that the initial velocity is negative.]
AnswerThe displacement rquation for an oscillating mass is given by:$\text{x}=\text{A}\cos(\omega\text{t}+\theta)$
where,
A is the amplitude
x is the displacement
$\theta$ is the phase constant
Velcoity, $\text{v}=\frac{\text{dx}}{\text{dt}}=-\text{A}\omega\sin(\omega\text{t}+\theta)$
At, $t = 0, x = x_0$
$\text{x}_0=\text{A}\cos\theta=\text{x}_0\ .....(\text{i})$
and, $\frac{\text{dx}}{\text{dt}}=-v_0=\text{A}\omega\sin\theta$
$\text{A}\sin\theta=\frac{v_0}{\omega}\ ....(\text{ii})$
Squaring and adding equations (i) and (ii), we get:
$\text{A}^2(\cos^2\theta+\sin^2\theta)=\text{x}_0^2+\Big(\frac{v_0^2}{\omega^2}\Big)$
$\therefore\ \text{A}=\sqrt{\text{x}_0^2+\Big(\frac{v_0}{\omega}\Big)^2}$
Hence, the amplitude of the resulting oscillation is $\text{x}_0^2+\Big(\frac{v_0}{\omega}\Big)^2.$
View full question & answer→Question 55 Marks
The motion of a particle executing simple harmonic motion is described by the displacement function,$\text{x(t)}=\text{A}\cos(\omega\text{t}+\phi).$
If the initial (t = 0) position of the particle is 1cm and its initial velocity is $\omega\text{ cm/s,}$ what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi\text{s}^{-1}.$ If instead of the cosine function, we choose the sine function to describe the SHM: $\text{x}=\text{B}\sin(\omega\text{t}+\alpha),$ what are the amplitude and initial phase of the particle with the above initial conditions.
Answer Initially, at t = 0: Displacement, x = 1cm Initial velocity, $\text{v}=\omega\text{ cm/sec.}$ Angular frequency, $\omega=\pi\text{ rad/s}^{-1}$ It is given that: $\text{x(t)}=\text{A}\cos(\omega\text{t}+\phi)$ $1=\text{A}\cos(\omega\times0\times\phi)$ $\text{A}\cos\phi=1\ ........(1)$ $\text{Velocity, }\upsilon=\frac{\text{dx}}{\text{dt}}$ $\omega=-\text{A}\omega\sin(\omega\text{t}+\phi)$ $1=-\text{A}\sin(\omega\times0+\phi)=-\text{A}\sin\phi$ $\text{A}\sin\phi=-1\ ....(2)$ Squaring and adding equations (1) and (2), we get: $\text{A}^2(\sin^2\phi+\cos^2\phi)=1+1$ $\text{A}^2=2$ $\therefore\ \text{A}=\sqrt{2}\ \text{cm}$ Dividing equation (2) by equation (1), we get: $\tan\phi=-1$ $\therefore\ \phi=\frac{3\pi}{4},\frac{7\pi}{4},....$ SHM is given as: $\text{x}=\text{B}\sin(\omega\text{t}+\alpha)$Putting the given values in this equation, we get:
$1=\text{B}\sin[\omega\times0+\alpha]$ $\text{B}\sin\alpha=1\ .....(3)$ Velocity, $\upsilon=\omega\text{B}\cos(\omega\text{t}+\alpha)$ Substituting the given values, we get: $\pi=\pi\text{B}\sin\alpha$ $\text{B}\sin\alpha=1\ .....(4)$ Squaring and adding equations (3) and (4), we get:$\text{B}^2[\sin^2\alpha+\cos^2\alpha]=1+1$
$\text{B}^2=2$
$\therefore\ \text{B}=\sqrt{2}\text{ cm}$
Dividing equation (3) by equation (4), we get: $\frac{\text{B}\sin\alpha}{\text{B}\cos\alpha}=\frac{1}{1}$ $\tan\alpha=1=\frac{\tan\alpha}{4}$ $\therefore\ \alpha=\frac{\pi}{4},\frac{5\pi}{4}, .....$
View full question & answer→Question 65 Marks
A spring balance has a scale that reads from $0$ to $50kg$. The length of the scale is $20cm$. A body suspended from this balance, when displaced and released, oscillates with a period of $0.6s$. What is the weight of the body?
AnswerMaximum mass that the scale can read, M = 50kg Maximum displacement of the spring = Length of the scale, l = 20cm = 0.2mTime period, T = 0.6s
Maximum force exerted on the spring, F = Mg
Where, g = acceleration due to gravity = $9.8m/s^2$
$F = 50 \times 9.8 = 490$
$\therefore$ Spring constant, $\text{k}=\frac{\text{F}}{\text{l}}=\frac{490}{0.2}=2450\text{ Nm}^{-1}$ Mass m, is suspended from the balance. Time period, $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
$\therefore\ \text{m}=\Big(\frac{\text{T}}{2\pi}\Big)^2\times\text{k}=\Big(\frac{0.6}{2\times3.14}\Big)^2\times2450$ = 22.36kg $\therefore$ Weight of the body = mg = 22.36 \times 9.8 = 219.167N Hence, the weight of the body is about 219N.
View full question & answer→Question 75 Marks
A spring having with a spring constant $1200N m^{-1}$ is mounted on a horizontal table as shown in Fig. A mass of $3kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0cm$ and released.
Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. AnswerSpring constant, $k = 1200N m^{-1}$
Mass, m = 3kg
Displacement, A = 2.0cm = 0.02cm
- Frequency of oscillation v, is given by the relation:
$\upsilon=\frac{1}{\text{T}}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
where, T is time period
$\therefore\ \upsilon=\frac{1}{2\times3.14}\sqrt{\frac{1200}{3}}$
= 3.18m/s
Hence, the frequency of oscillations is 3.18 cycles per second.
- Maximum acceleration (a) is given by the relation:
$\text{a}=\omega^2\text{A}$
where,
$\omega=$ Angular frequency $=\sqrt{\frac{\text{k}}{\text{m}}}$
A = maximum displacement
$\therefore\ \text{a}=\frac{\text{k}}{\text{m}}\text{A}=\frac{1200\times0.02}{3}=8\text{ ms}^{-2}$
Hence, the maximum acceleration of the mass is $8.0m/s^2$.
Maximum velocity, $\text{v}_\text{max}=\text{A}\omega$
$=\text{A}\sqrt{\frac{\text{k}}{\text{m}}}=0.02\times\sqrt{\frac{1200}{3}}=0.4\text{ m/s}$
Hence, the maximum velocity of the mass is 0.4m/s. View full question & answer→Question 85 Marks
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F.
If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case?
Answer In Fig. (a) if x is the extension in the spring, when mass m is returning to its mean position after being released free, then restoring force on the mass iss F = -kx, i.e., $\text{F}\propto\text{x}$
As, this F is directed towards mean position of the mass, hence the mass attached to the spring will execute SHM.
Spring factor = spring constant = k
inertia factor = mass of the given mass = m
As time period,
$\text{T}=2\pi\sqrt{\frac{\text{inertia factor}}{\text{spring factor}}}$
$\therefore\ \text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
In Fig. (b), we have a two body system of spring constant k and reduced mass, $\mu=\frac{\text{m}\times\text{m}}{\text{m+m}}=\frac{\text{m}}{2}$
Inertia factor = m/2
Spring factor = k
$\therefore$ Time period, $\text{T}=2\pi\sqrt{\frac{\frac{\text{m}}{2}}{\text{k}}}=2\pi\sqrt{\frac{\text{m}}{2\text{k}}}$
View full question & answer→Question 95 Marks
A circular disc of mass $10kg$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be $1.5s$. The radius of the disc is 15cm. Determine the torsional spring constant of the wire. (Torsional spring constant $\alpha$ is defined by the relation $\text{J}=-\alpha\theta,$ where J is the restoring couple and $\theta$ the angle of twist).
AnswerMass of the circular disc, m = 10kg
Radius of the disc, r = 15cm = 0.15m
The torsional oscillations of the disc has a time period, T = 1.5s
The moment of inertia of the disc is:
$\text{l}=\frac{1}{2}\text{mr}^2$
$=\frac{1}{2}\times(10)\times(0.15)^2$
$= 0.1125kg m^2$
Time period, $\text{T}=2\pi\sqrt{\frac{\text{l}}{\alpha}}$
$\alpha$ is the torsional constant.
$\alpha=\frac{4\pi^2\text{l}}{\text{T}^2}$
$=\frac{4\times(\pi)^2\times0.1125}{(1.5)^2}$
= 1.972Nm/rad
Hence, the torsional spring constant of the wire is $1.972Nm\ rad^{-1}$.
View full question & answer→Question 105 Marks
A body describes simple harmonic motion with an amplitude of 5cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is (a) 5cm (b) 3cm (c) 0cm.
Answer Amplitude, A = 5cm = 0.05m Time period, T = 0.2s For displacement, x = 5cm = 0.05m Acceleration is given by: $\text{a}=-\omega^2\text{x}$ $=-\Big(\frac{2\pi}{\text{T}}\Big)^2\text{x}$ $=-\Big(\frac{2\pi}{0.2}\Big)^2\times0.05$ $=-5\pi^2\text{ m/s}^2$ Velocity is given by: $\text{v}=\omega\sqrt{\text{A}^2-\text{x}^2}$ $=\frac{2\pi}{\text{T}}\sqrt{(0.05)^2-(0.05)^2}$ = 0When the displacement of the body is 5cm, its acceleration is $-5\pi^2\text{ m/s}^2$ and velocity is 0.
For displacement, x = 3cm = 0.03m
$\text{a}=-\omega^2\text{x}$ $=-\Big(\frac{2\pi}{\text{T}}\Big)^2\text{x}$ $=-\Big(\frac{2\pi}{0.2}\Big)^20.03$$=-3\pi^2\text{ m/s}^2$
Velocity is given by: $\text{v}=\omega\sqrt{\text{A}^2-\text{x}^2}$ $=\frac{2\pi}{\text{T}}\sqrt{\text{A}^2-\text{x}^2}$$=\frac{2\pi}{\text{T}}\sqrt{(0.05)^2-(0.03)^2}$
$=\frac{2\pi}{0.2}\times0.04$
$=0.4\pi\text{ m/s}$
When the displacement of the body is 3cm, its acceleration is $-3\pi\text{ m/s}^2$ and velocity is $0.4\pi\text{ m/s}.$ For displacement, x = 0 Acceleration is given by: $\text{a}=-\omega^2\text{x}=0$ Velocity is given by: $\text{v}=\omega\sqrt{\text{A}^2-\text{x}^2}$ $=\frac{2\pi}{\text{T}}\sqrt{\text{A}^2-\text{x}^2}$ $=\frac{2\pi}{0.2}\sqrt{(0.05)^2-0}$$=0.5\pi\text{ m/s}$
When the displacement of the body is 0, its acceleration is 0 and velocity is $0.5\pi\text{ m/s}.$
View full question & answer→Question 115 Marks
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Answer The equation of displacement of a particle executing SHM at an instant t is given as:$\text{x}=\text{A}\sin\omega\text{t}$
where,
A = Amplitude of oscillation
$\omega=$ Angular frequency $=\sqrt{\frac{\text{k}}{\text{M}}}$
The velocity of the particle is: v = dx/dt $=\text{A}\omega\cos\omega\text{t}$
The kinetic energy of the particle is:
$\text{E}_\text{k}=\frac{1}{2}\text{Mv}^2=\frac{1}{2}\text{MA}^2\omega^2\cos^2\omega\text{t}$
The portential energy of the particle is:
$\text{E}_\text{p}=\frac{1}{2}\text{kx}^2=\frac{1}{2}\text{M}^2\omega^2\text{A}^2\sin^2\omega\text{t}$
For time period T, the average kinetic energy over a single cycle is given as:
$(\text{E}_\text{k})_\text{Avg}=\frac{1}{\text{T}}\int\limits_{0}^{\text{T}}\text{E}_\text{k}\text{dt}$
$=\frac{1}{\text{T}}\int\limits_{0}^{\text{T}}\frac{1}{2}\text{MA}^2\omega^2\cos^2\omega\text{t dt}$
$=\frac{1}{2\text{T}}\text{MA}^2\omega^2\int\limits_{0}^{\text{T}}\frac{(1+\cos2\omega\text{t})}{2}\text{dt}$
$=\frac{1}{4\text{T}}\text{MA}^2\omega^2\Big[\text{t}+\frac{\sin2\omega\text{t}}{2\omega}\Big]^{\text{T}}_{0}$
$=\frac{1}{4\text{T}}\text{MA}^2\omega^2(\text{T})$
$=\frac{1}{4}\text{MA}^2\omega^2\ ......(\text{i})$
And, average potential energy over one cycle is given as:
$(\text{E}_\text{p})_\text{Avg}=\frac{1}{\text{T}}\int\limits^{\text{T}}_{0}\text{E}_\text{p}\text{dt}$
$=\frac{1}{\text{T}}\int\limits_{0}^{\text{T}}\frac{1}{2}\text{MA}^2\omega^2\sin^2\omega\text{t dt}$
$=\frac{1}{2\text{T}}\text{MA}^2\omega^2\int\limits_{0}^{\text{T}}\frac{(1-\cos2\omega\text{t})}{2}\text{dt}$
$=\frac{1}{4\text{T}}\text{MA}^2\omega^2\Big[\text{t}-\frac{\sin2\omega\text{t}}{2\omega}\Big]^{\text{T}}_{0}$
$=\frac{1}{4\text{T}}\text{MA}^2\omega^2(\text{T})$
$=\frac{1}{4}\text{MA}^2\omega^2\ ......(\text{ii})$
It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period.
View full question & answer→Question 125 Marks
You are riding in an automobile of mass $3000\ kg$. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by $50\%$ during one complete oscillation. Estimate the values of $(a)$ the spring constant $k$ and $(b)$ the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports $750\ kg.$
Answer
- Mass of the automobile, $m = 3000kg$
Displacement in the suspension system, $x = 15cm = 0.15m$
There are $4$ springs in parallel to the support of the mass of the automobile.
The equation for the restoring force for the system:
$F = –4kx = mg$
Where, $k$ is the spring constant of the suspension system
Time period, $\text{T}=2\pi\sqrt{\frac{\text{m}}{4\text{k}}}$
and $k = mg/4x = 3000 \times 10/4 \times 0.15 = 5000 = 5 \times 10^4Nm$
Spring Constant, $k = 5 \times 10^4Nm$
- Each wheel supports a mass, $M = 3000/4 = 750kg$
For damping factor $b$, the equation for displacement is written as
$x = x_0e^{-bt/2M}$
The amplitude of oscilliation decreases by $50\%.$
$\therefore x = x_0/2$
$x_0/2 = x_0e^{-bt/2M}$
$\log_e2 = bt/2M$
$\therefore b = 2M log_e2/t$
where,
Time period, $\text{t}=2\pi\sqrt{\frac{\text{m}}{4\text{k}}}=2\pi\sqrt{\frac{3000}{4\times5\times10^4}}=0.7691\text{ s}$
$\therefore\ \text{b}=\frac{2\times750\times0.693}{0.7691}=1351.58\text{ kg/s}$
View full question & answer→Question 135 Marks
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density $\rho_1.$ The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period $\text{T}=2\pi\sqrt{\frac{\text{h}\rho}{\rho_1\text{g}}}$ where $\rho$ is the density of cork. (Ignore damping due to viscosity of the liquid).
Answer Base area of the cork = A Height of the cork = h Density of the liquid = $\rho_1$ Density of the cork = $\rho$ In equilibrium: Weight of the cork = Weight of the liquid displaced by the floating cork Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork. Up-thrust = Restoring force, F = Weight of the extra water displaced F = –(Volume × Density × g) Volume = Area × Distance through which the cork is depressed Volume = Ax$\therefore\ \text{F}=-\text{Ax}\rho_1\text{g}\ .....(\text{i})$
Accroding to the force law: F = kx k = F/x where, k is constant$\text{k}=\frac{\text{F}}{\text{x}}=-\text{A}\rho_1\text{g}\ .....(\text{ii})$
The time period of the oscillations of the cork:$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}\ .....(\text{iii})$
where, m = Mass of the cork = Volume of the cork × Density = Base area of the cork × Height of the cork × Density of the cork$=\text{Ah}\rho$
Hence, the expression for the time period becomes: $\text{T}=2\pi\sqrt{\frac{\text{Ah}\rho}{\text{A}\rho_1\text{g}}}=2\pi\sqrt{\frac{\text{h}\rho}{\rho_1\text{g}}}$
View full question & answer→Question 145 Marks
An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.). Show that when the ball is pressed down a little and released , it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig.].
Answer Volume of the air chamber = V Area of cross-section of the neck = a Mass of the ball = m The pressure inside the chamber is equal to the atmospheric pressure. Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber Decrease in the volume of the air chamber, ΔV = ax Volumetric strain = Change in volume/Original volume ⇒ ΔV/V = ax/V Bulk Modulus of air B = Stress/Strain = -p/ax/V In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume. $\text{p}=\frac{-\text{Bax}}{\text{V}}$ The restoring force acting on the ball, $\text{F}=\text{p}\times\text{a}$ $=\frac{-\text{Bax}}{\text{V}}.\text{a}$ $=\frac{-\text{Ba}^2\text{x}}{\text{V}}\ ....(\text{i})$ In simple harmonic motion, the equation for restoring force is:F = -kx ....(ii)
Where, k is the spring constant Comparing equations (i) and (ii), we get: $\text{k}=\frac{\text{Ba}^2}{\text{V}}$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$ $=2\pi\sqrt{\frac{\text{Vm}}{\text{Ba}^2}}$
View full question & answer→Question 155 Marks
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Answer Area of cross-section of the U-tube = A Density of the mercury column = $\rho$ Acceleration due to gravity = g Restoring force, F = Weight of the mercury column of a certain height F = –(Volume × Density × g) $\text{F}=-(\text{A}\times2\text{h}\times\rho\times\text{g})=-2\text{A}\rho\text{gh}$= -k × Displacement in one of the arms (h)
Where, 2h is the height of the mercury column in the two arms k is a constant, given by $\text{k}=-\frac{\text{F}}{\text{h}}=2\text{A}\rho\text{g}$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{\text{m}}{2\text{A}\rho\text{g}}}$ Where, m is the mass of the mercury column Let l be the length of the total mercury in the U-tube. Mass of mercury, m = Volume of mercury × Density of mercury $=\text{Al}\rho$ $\therefore\ \text{T}=2\pi\sqrt{\frac{\text{Al}\rho}{2\text{A}\rho\text{g}}}=2\pi\sqrt{\frac{\text{l}}{2}}\text{g}$ Hence, the mercury column executes simple harmonic motion with time period $2\pi\sqrt{\frac{\text{l}}{2\text{g}}}$
View full question & answer→Question 165 Marks
Two linear simple harmonic motions of equal amplitudes and frequencies $\omega$ and $2\omega$ are impressed on a particle along the axes of X and Y respectively. If the initial phase difference between them $\frac{\pi}{2}$ is find the resultant path followed by the particle.
Answer Two simple harmonic motions of equal amplitudes (A) and frequencies o and 20 and initial phase difference of $\frac{\pi}{2}$ are represented by$\text{x}=\text{A}\sin\omega\text{t}\cdots\text{(i)}$
$\text{y}=\text{A}\sin\Big(2\omega\text{t}+\frac{\pi}{2}\Big)$
$=\text{A}\cos2\omega\text{t}\cdots\text{(ii)}$
Since $\cos2\omega\text{t}=(1-2\sin^2\omega\text{t})$$\therefore\text{y}=\text{A}[1-2\sin^2\omega\text{t}]\cdots\text{(iii)}$
From equ.(i), $\sin^2\omega\text{t}=\frac{\text{x}^2}{\text{A}^2}$$\therefore\text{y}=\text{A}\Big[1-\frac{2\text{x}^2}{\text{A}^2}\Big]$
$=\text{A}-\frac{2\text{x}^2}{\text{A}}$
$\Rightarrow\frac{2\text{x}^2}{\text{A}}+\text{y}-\text{A}=0$
$=\text{x}^2+\frac{\text{Ay}}{2}-\frac{\text{A}^2}{2}=0$
Which is the equation of a parabola. Hence the resultant path followed by the particle is parabolic.
View full question & answer→Question 175 Marks
A particle performs SHM on a rectilinear path. Starting from rest, it travels $x_1$ distance in first second and in the next second, it travels $x_2$ distance. Find out the amplitude of this SHM.
AnswerBecause the particle starts from rest, so its starting point will be extreme position. Thus, the displacement of the particle from the mean position after one second$\text{A}-\text{x}_1=\text{A}\cos\omega\cdots\text{(i)}$ [[putting t = 1s]
$\text{A}-(\text{x}_1+\text{x}_2)$
$\text{A}\Big[2\Big(\frac{\text{A}-\text{x}_1}{\text{A}}\Big)^2-1\Big]$
$=\frac{1}{\text{A}}[2\text{A}^2+2\text{x}^2_1-4\text{Ax}_1-\text{A}]$
$\Rightarrow\text{A}^2-\text{A}(\text{x}_1+\text{x}_2)$
$=\text{A}^2+2\text{x}_1^2-4\text{Ax}_1$
$\Rightarrow\text{A}[3\text{x}_1-\text{x}_2]=2\text{x}_1^2$
$\therefore\text{A}=\frac{2\text{x}_1^2}{3\text{x}_1-\text{x}_2}$
View full question & answer→Question 185 Marks
A spring compressed by 0.1m develops a restoring force 10N. A body of mass 4kg placed on it. Deduce
- The force constant of the spring.
- The depression of the spring under the weight of the body(take g = 10 N/ kg).
- The period of oscillation, the body is distributed and.
- Frequency of oscillation.
Answer Here,$\text{F}=10\text{N}$$\Delta\text{l}=0.1\text{m}$
$\text{m}=4\text{kg}$
- $\text{k}=\frac{\text{F}}{\Delta\text{l}}$
$=\frac{10}{0.1}=100\text{Nm}^{-1}$
- $\text{y}=\frac{\text{mg}}{\text{k}}$
$=\frac{4\times10}{100}=0.4\text{m}$
- $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
$=2\times\frac{22}{7}\sqrt{\frac{4}{100}}$
$=1.26\text{s}$
- Frequency. $\text{v}=\frac{1}{\text{T}}$
$=\frac{1}{1.26}=0.8\text{Hz}$ View full question & answer→Question 195 Marks
An object of mass 0.2kg executes simple harmonic oscillations along the x-axis with a frequency of $\frac{25}{\pi}$ Hertz. At the position x = 0.04m, the object has kinetic energy of 0.5J and potential energy of 0.4J. Find the amplitude of oscillations.
Answer Given that, $\text{m}=0.2\text{kg},\text{v}=\frac{25}{\pi}\text{Hz}$$\therefore\omega=2\pi\text{v}=2\pi\times\frac{25}{\pi}$
$=50\text{ rad s}^{-1}$
$\text{x}=0.04\text{m, E}_{\text{K}}=0.5\text{J},$
$\text{E}_{\text{p}}=0.4\text{J}$
We know,$\text{E}_{\text{K}}=\frac{1}{2}\text{m}\omega^2(\text{a}^2-\text{x}^2)$
and $\text{E}_{\text{P}}=\frac{1}{2}\text{m}\omega^2\text{x}^2$ Dividing the two,$\frac{\text{E}_{\text{K}}}{\text{E}_{\text{P}}}=\frac{\text{a}^2-\text{x}^2}{\text{x}^2}=\frac{\text{a}^2}{\text{x}^2}-1$
$\frac{\text{a}^2}{\text{x}^2}=\frac{\text{E}_{\text{K}}}{\text{E}_{\text{P}}}+1$
$\text{a}^2=\Big(\frac{\text{E}_{\text{K}}}{\text{E}_{\text{P}}}+1\Big)\text{x}^2$
$=\Big(\frac{0.5}{0.4}+1\Big)(0.04)^2$
$=\frac{9}{4}\times0.04\times0.04$
$=36\times10^{-4}\text{m}^2$
$\text{a}=\sqrt{36\times10^{-4}}\text{m}=6\times10^{-2}\text{m}$
View full question & answer→Question 205 Marks
The mass 'M' attached to a spring oscillates with a period 2s. If the mass is increased by 2kg, the period increases by 1s. Find the initial mass 'M', assuming that Hooke's law is obeyed.
Answer Let the initial mass and time periods be M and T respectively. If Hooke's law is obeyed, then the oscillations of the spring will be simple harmonic having time period T given by $\text{T}=2\pi\sqrt{\frac{\text{M}}{\text{k}}}$ Given T = 2s$\therefore2\pi=2\pi\sqrt{\frac{\text{M}}{\text{k}}}$ k = spring constant ...(i)
On increasing the mass by 2kg$3=2\pi\sqrt{\frac{\text{M}+2}{\text{k}}}\cdots\text{(ii)}$
Squaring and dividing equation (ii) by (i) we have$\frac{9}{4}=\frac{\text{M}+2}{\text{M}}=1+\frac{2}{\text{M}}$
$=\frac{2}{\text{M}}=\frac{9}{4}-1=\frac{5}{4}$
$\therefore\text{M}=\frac{2\times4}{5}$
$=\frac{8}{5}=1.6\text{kg}$
View full question & answer→Question 215 Marks
Suppose a tunnel is dug through the earth from one side to the other side along a diameter. Show that the motion of a particle dropped into the tunnel is simple harmonic motion. Find the time period. Neglect all the frictional forces and assume that the earth has a uniform density.
$G = 6.67 \times 10^{-11}Nm^2kg^{-2};$ density of earth = $5.51 \times 10^3kg m^{-3}$
AnswerFigure shows a tunnel dug along the diameter of the earth. Consider the case of a particle of mass m at a distance y from the centre of the earth. There will be a gravitational attraction of the earth on this particle due to the portion of matter contained in a sphere of radius y. The mass of the sphere of radius y is given by
M = Volume × density$\text{M}=\frac{4}{3}\pi\text{y}^3\times\text{d}$
This mass can be regarded as concentrated at the centre of the earth. The force F between this mass and the particle of mass m is given by$\text{F}=-\frac{\text{GMm}}{\text{y}^2}$
Negative sign shows that the force is of attraction.$\therefore\text{F}=-\text{G}\Big(\frac{4}{3}\pi\text{y}^3\text{d}\Big)\frac{\text{m}}{\text{y}}^2$
$-\text{G}\Big(\frac{4}{3}\pi\text{m}\text{d}\Big)\text{y}$
$\text{F}\propto\text{y}$
The force is directly proportional to the displacement, hence the motion is simple harmonic motion. Here, the constant $\text{k}\frac{4}{3}\pi\text{mdG}$. The time period, $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$$=2\pi\sqrt{\Big(\frac{3\text{m}}{4\pi\text{mdG}}\Big)}$
$=2\pi\sqrt{\Big(\frac{3\text{m}}{4\pi\text{dG}}\Big)}$
$=\sqrt{\Big(\frac{3\pi}{\text{dG}}\Big)}$
$=\sqrt{\Big(\frac{3\times3.14}{5.51\times10^3\times6.67\times10^{-11}}\Big)}$
$=42.2\text{ minutes}$. View full question & answer→Question 225 Marks
What is the frequency of a second pendulum in an elevator moving up with an acceleration of $\frac{\text{g}}{2}$?
Answer For second pendulum, frequency $\text{v}=\frac{1}{2}\text{s}^{-1}$ When elevator is moving upwards with acceleration a, the effective acceleration due to gravity is $\text{g}_1=\text{g}+\text{a}=\text{g}+\frac{\text{g}}{2}=\frac{3\text{g}}{2}$ Since $\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{g}}{\text{l}}}$ Hence, $\text{v}^2\propto\text{g}$$\therefore\frac{\text{v}_1}{\text{v}_2}=\frac{\text{g}_1}{\text{g}}$
$\frac{\frac{3\text{g}}{2}}{\text{g}}=\frac{3}{2}$
$=\frac{\text{v}_1}{\text{v}}=\sqrt{\frac{3}{2}}$
$=1.225$
$\Rightarrow\text{v}_1=1.225$
$=\text{v}=1.225\times\frac{1}{2}$
$=0.612\text{s}^{-1}$
View full question & answer→Question 235 Marks
A trolley of mass $3.0kg$, as shown in Figure, is connected to two springs, each of spring constant $600Nm^{-1}$. If the trolley is displaced from its equilibrium position by $5.0cm$ and released, what is (a) the period of ensuing oscillations, and (b) the maximum speed of the trolley? How much energy is dissipated as heat by the time trolley comes to rest due damping forces?
AnswerEquivalent spring conslant = $k' = 2k = 1200Nm^{-1}, m = 3kg$
$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{3}{1200}}$
$=\frac{2\pi}{20}=\frac{\pi}{10}\text{sec}.$
Maximum speed$\text{v}=\omega\text{A}=20\times5\times-10^{-2}=1\text{ms}^{-1}$
Energy dissipated = Maximum energy$=\frac{1}{2}\text{m}\omega^2\text{A}^2$
$=\frac{1}{2}\times3\times20^2\times25\times10^{-4}$
$=600\times25\times10^{-4}$
$=150\times10^{-2}\text{ Joule}$
$=1.5\text{ Joule}$
View full question & answer→Question 245 Marks
The periodic time of a mass suspended by a spring (force constant K) is T. If the spring is cut in three equal pieces, what will be the force constant of each part? If the same mass be suspended from one piece, what will be the periodic time?
Answer Consider the spring be made of a combination of three springs in series each of spring constant k. The effective spring constant K is given by $\frac{1}{\text{K}}=\frac{1}{\text{k}}+\frac{1}{\text{k}}+\frac{1}{\text{k}}=\frac{3}{\text{k}}$ $=\text{K}=\frac{\text{k}}{3}$
$=\text{k}=3\text{k}$
$\therefore$ Time period of vibration of a body attached to the end of this spring,
$\text{T}=2\text{k}\sqrt{\frac{\text{m}}{\text{k}}}$
$=2\pi\sqrt{\frac{\text{m}}{\big(\frac{\text{k}}{3}\big)}}$
$=2\pi\sqrt{\frac{3\text{m}}{\text{k}}}\cdots\text{(i)}$
When the spring is cut into three pieces the spring constant= k time period of vibration of a body attached to the end of this spring,$\text{T}_1=2\pi\sqrt{\frac{\text{m}}{\text{k}}}\cdots\text{(ii)}$
From eqns.(i) and (ii),$\frac{\text{T}_1}{\text{T}}=\frac{1}{\sqrt{3}}$
$=\text{T}_1=\frac{\text{T}_1}{\sqrt{3}} $
View full question & answer→Question 255 Marks
A block whose mass is $1kg$ is fastened to a spring. The spring has a spring constant of $50Nm^{-1}$. The block is pulled to a distance $x = 10cm$ from the equilibrium position at $x = 0$ on a frictionless surface from rest at $t = 0$. Calculate the kinetic, potential and total energies of the block when it is $5cm$ away from the mean position.
AnswerHere mass of block m = 1 kg and spring constant $k = 50 Nm^{-1}$
$\therefore$ Angular frequency of SHM $\omega=\sqrt{\frac{\text{k}}{\text{m}}}$
$=\sqrt{\frac{50}{1}}$
$=7.07\text{ rad s}^{-1}$
As at t = 0, the block was x = 0, and then the block was pulled to distance x = 10cm, it shows that amplitude of oscillation A = 10 - 0 = 10cm = 0.1m. Moreover at any instant the instantaneous displacement x(t) = 5cm = 0.005m$\therefore$ Kinetic energy of the block $\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$
$\frac{1}{2}\times1\times(7.07)^2[(0.05)]$
$=0.19\text{J}$
Potential energy of the block $\text{U}=\frac{1}{3}\text{m}\omega^2\text{x}^2$$=\frac{1}{2}\times1\times(7.07)^2\times(0.05)^2$
$=6.25\times10^{-2}\text{J}$
Total energy of thr block $\text{E}=\text{K}+\text{U}$$=0.19\text{J}+6.25\times10^{-2}\text{J}$
$=0.25\text{J}$
View full question & answer→Question 265 Marks
Two pendulums of lengths 100cm and 110.25cm start oscillating in phase simultaneously. After how many oscillations will they again be in phase together?
Answer $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$$\text{l}_1=100\text{cm}$
$\text{l}_2=110.25\text{cm}$
For smaller pendulum, $\text{T}_1=2\pi\sqrt{\frac{100}{\text{g}}}\cdots\text{(i)}$
For larger pendulum, $\text{T}_2=2\pi\sqrt{\frac{110.25}{\text{g}}}\cdots\text{(ii)}$
Let these pendulums oscillate in phase again if larger pendulum completes 'n' oscillations. It means smaller pendulum must complete (n + 1) oscillations.
$\text{nT}_2=(\text{n}+1)\text{T}_1$
$\frac{\text{n}+1}{\text{n}}$
$\frac{\text{T}_2}{\text{T}_1}=\sqrt{\frac{110.25}{100}}$
$=1.05$
$=1+\frac{1}{\text{n}}$
$=1.05$
$=\frac{1}{\text{n}}=0.05$
$=\frac{5}{100}=\frac{1}{20}$
$\therefore\text{n}=20$
Hence both pendulums will again oscillate in phase after 20 oscillations of the larger or 21 oscillations of the smaller pendulum.
View full question & answer→Question 275 Marks
A simple pendulum with a brass has a time period T. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is $\frac{1}{9}$ that of brass, find the time period of the same pendulum.
Answer Let V be the volume and P be the density of the brass bob. Mass of the bob $\text{m}=\text{V}\rho$ and weight of bob $\text{V}\rho\text{g}$. Buoyancy force of liquid on bob $=\text{V}\frac{\rho}{9}\text{g}$$=\frac{\text{V}\rho\text{g}}{9}$
So the effective weight of bob in liquid $\text{V}\rho\text{g}-\frac{\text{V}\rho\text{g}}{9}$$=8\frac{\text{V}\rho\text{g}}{9}$
$\therefore$ Acceleration $\text{g}'=\frac{\frac{8\text{V}\rho\text{g}}{9}}{\text{m}}$$=\frac{\frac{8\text{V}\rho\text{g}}{9}}{\text{V}\rho}$
$=\frac{8\text{g}}{9}$
Time period of the bob $=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$$=2\pi\sqrt{\frac{\text{l}}{\frac{8\text{g}}{9}}}$
$=2\pi\sqrt{\frac{\text{l}}{\text{g}}}\times\frac{3}{\sqrt{8}}$
$=\frac{3\text{T}}{\sqrt{8}}$
View full question & answer→Question 285 Marks
Figures correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
AnswerLet B be any point on the circle of reference of the figure. From B draw BN perpendicular on x-axis. Then $\triangle\text{BON}=\theta=\omega\text{t}$$\therefore$ In $\triangle\text{ONB},\ \cos\theta=\frac{\text{ON}}{\text{OB}}$
Or $\text{ON}=\text{OB}\cos\theta$$\therefore\ -\text{x}=2\cos\omega\text{t}$
$\Rightarrow\ \text{x}=-2\frac{\cos(2\pi)}{\text{T}}\text{t}=-2\cos\frac{2\pi}{4}\text{t}$
$\therefore\ \text{x}=-2\cos\frac{\pi}{4}\text{t}$ which is equation of SHM
View full question & answer→Question 295 Marks
A uniform spring whose unstretched length is l has a force constant k. The spring is cut into two pieces of unstretched lengths. $l_1$, and $l_2$ where $l_1 = nl_2$, and n is an integer. What are the corresponding force constants $k_1$ and $k_1$ in terms of n and k?
AnswerHere $l = l_1 + l_2$ ...(i) and $l_1 = nl_2$ ...(ii) We khow, $\text{k}=\frac{\text{Mg}}{\text{l}}\cdots\text{(iii)}$$\therefore\text{k}_1=\frac{\text{Mg}}{\text{l}_1}\cdots\text{(iv)}$
$\text{k}_2=\frac{\text{Mg}}{\text{l}_2}\cdots\text{(v)}$
Dividing equation (iv) by equation (iii) we find$\frac{\text{k}_1}{\text{k}}=\frac{\text{l}}{\text{l}_1}=\frac{\text{l}_1+\text{l}_2}{\text{l}_1}=1+\frac{\text{l}_2}{\text{l}_1}$
From equation (ii) we find$\frac{\text{l}_1}{\text{l}_2}=\text{n}$
$\therefore\frac{\text{k}_1}{\text{k}}=1+\frac{1}{\text{n}}$
$\text{k}_1=\Big(\frac{\text{n}+1}{\text{n}}\Big)\text{k}$ From equation (ii) and (iii), we find:$\frac{\text{k}_2}{\text{k}}=\frac{\text{l}}{\text{l}_2}=\frac{\text{l}_1+\text{l}_2}{\text{l}_2}=\frac{\text{l}_1}{\text{l}_2}+1$
From equation (ii) we have $\frac{\text{l}_1}{\text{l}_2}=\text{n}$.$\therefore\frac{\text{k}_2}{\text{k}}=(\text{n}+1)$
$\therefore\text{k}_2=\text{k}(\text{n}+1)$
View full question & answer→Question 305 Marks
A particle executes S.H.M of time period $10s$. The displacement at any instant is given by the relation $\text{x}=10\sin\omega\text{t}$ Find
- Velocity of the body 2s after it passes through the mean position and
- The acceleration 2s after it passes the mean position (Amplitude is given in cm).
Answer
- Velocity at any instant t is given by $\text{v}=\text{A}\omega\cos\text{t}$
Here A = 10cm
$\omega=\frac{2\pi}{\text{T}}=\frac{2\pi}{10}$
When t = 2s,
$\text{v}=10\times\frac{2\pi}{10}\cos\Big(\frac{2\pi}{10}\times2\Big)$
$=2\pi\cos(0.4\pi)$
$=1.942\text{cm/ s}$
- Accleration at any instant t is given by
$\text{a}=-\text{A}\omega^2\sin\omega\text{t}$
$=-10\big(\frac{2\pi}{10}\big)^2\sin(0.4\pi)$
$=-3.755\text{cm/ s}^2$
Acceleration is numerically equal to $3.754cm/ s^2$ and is directed towards the mean position. View full question & answer→Question 315 Marks
Find the total energy of the particle executing S.H.M. and show graphically the variation of P.E. and K.E. with time in S.H.M. What is the frequency of these energies with respect to the frequency of the particle executing S.H.M.
Answerln a S.H.M.. with y $=\text{A}\sin\omega\text{t},$$\text{P.E.}=\frac{1}{2}\text{ky}^2=\frac{1}{2}\text{m}\omega^2\text{y}^2$
$=\frac{1}2{}\text{m}\omega^2\text{A}^2\sin^2\text{t}$
$\text{K.E.}=\frac{1}{2}\text{mv}^2=\frac{1}{2}\text{m}\Big(\frac{\text{dy}}{\text{dt}}\Big)^2$
$=\frac{1}{2}\text{m}\omega^2\text{A}^2\cos^2\omega\text{t}$
Total energy $=\text{P.E. + K.E.}$$=\frac{1}{2}\text{m}\omega^2\text{A}^2$
as $\sin^2\omega\text{t}+\cos^2\omega\text{t}=1$
Since, both P.E. and K.E. are square sinusoidal functions, their frequency will be double of a normal simple harmonic function. So, P.E. and K.E. will have a frequency 2f for a S.H.M. with frequency f. Total energy $=\frac{1}{2}\text{m}\omega^2\text{A}^2$ is a constant and so, there is no variation. View full question & answer→Question 325 Marks
What is the ratio between the distance travelled by the oscillator in one time period and amplitude?
AnswerIn the diagram shown a particle is executing SHM between P and Q. The particle starts from mean position ‘O’ moves to amplitude position ‘P’, then particle turn back and moves from ‘P’ to iQ Finally the particle turns back again and return to mean position ‘O’. In this way the particle completes one oscillation in one time period.
Total distance travelled while it goes from O → P → O → Q → O
= OP + PO + OQ + QO = A + A + A + A = 4A
Amplitude = OP = A
Hence, ratio of distance and amplitude = 4A/ A = 4.
View full question & answer→Question 335 Marks
The acceleration due to gravity on the surface of moon is $1.7m s^{-2}$. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is $3.5$ s? (g on the surface of earth is $9.8m s^{-2}$)
AnswerAcceleration due to gravity on the surface of moon, $g' = 1.7m s^{–2}$ Acceleration due to gravity on the surface of earth, $g = 9.8m s^{–2}$ Time period of a simple pendulum on earth, T = 3.5s$\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$
Where l is the length of the pendulum$\therefore\ \text{l}=\frac{\text{T}^2}{(2\pi)^2}\times\text{g}$
$=\frac{(3.5)^2}{4\times(3.14)^2}\times9.8\text{m}$
The length of the pendulum remains constant. On moon's surface, time period, $\text{T}'=2\pi\sqrt{\frac{\text{l}}{\text{g}'}}$$=2\pi\sqrt{\frac{\frac{(3.5)^2}{(4\times3.14)^2}\times9.8}{1.7}}=8.4\text{s}$
Hence, the time period of the simple pendulum on the surface of moon is 8.4s.
View full question & answer→Question 345 Marks
Given the example of the motion in the following cases:
- Where magnitude and direction of the acceleration of the particle changes.
- Where the magnitude and direction of acceleration of body remains constant.
- Where magnitude of acceleration changes but its direction remains constants.
- Where the magnitude of acceleration remains constant but its direction changes.
Answer
- In S.H.M., acceleration is always proportional to displacement but directed opposite to the displacement. So in this case, magnitude as well as direction of acceleration changes.
- A body falling under gravity near the surface of the earth.
- A body falling under gravity from a height comparable to the radius of the earth.
- A body revolving in a circular path with constant speed.
View full question & answer→Question 355 Marks
A mass attached to a spring is free to oscillate, with angular velocity $\omega,$ in a horizontal plane without friction or damping. It is pulled to a distance $x_0$ and pushed towards the centre with a velocity $\upsilon_0$ at time $t = 0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_0$ and $\upsilon_0.$ [Hint: Start with the equation $\text{x}=\text{a}\cos(\omega\text{t}+\theta)$ and note that the initial velocity is negative.]
AnswerThe displacement rquation for an oscillating mass is given by:$\text{x}=\text{A}\cos(\omega\text{t}+\theta)$
where,
A is the amplitude
x is the displacement
$\theta$ is the phase constant
Velcoity, $\text{v}=\frac{\text{dx}}{\text{dt}}=-\text{A}\omega\sin(\omega\text{t}+\theta)$
At, $t = 0, x = x_0$
$\text{x}_0=\text{A}\cos\theta=\text{x}_0\ .....(\text{i})$
and, $\frac{\text{dx}}{\text{dt}}=-v_0=\text{A}\omega\sin\theta$
$\text{A}\sin\theta=\frac{v_0}{\omega}\ ....(\text{ii})$
Squaring and adding equations (i) and (ii), we get:
$\text{A}^2(\cos^2\theta+\sin^2\theta)=\text{x}_0^2+\Big(\frac{v_0^2}{\omega^2}\Big)$
$\therefore\ \text{A}=\sqrt{\text{x}_0^2+\Big(\frac{v_0}{\omega}\Big)^2}$
Hence, the amplitude of the resulting oscillation is $\text{x}_0^2+\Big(\frac{v_0}{\omega}\Big)^2.$
View full question & answer→Question 365 Marks
A pendulum clock gives correct time. What is the error in time per day if the length increases by 0.05%?
Answer$\text{T}=2\pi\sqrt{\frac{\text{L}}{\text{g}}}$ and $\text{T}'=2\pi\sqrt{\frac{\big(\text{L}+\frac{0.05}{100}\text{L}\big)}{\text{g}}}$Dividing, $\frac{\text{T}'}{\text{T}}=\sqrt{\frac{\big(\text{L}+\frac{0.05}{100}\text{L}\big)}{\text{g}}}$
$=\Big[1+\frac{0.05}{100}\Big]^{\frac{1}2{}}$
Applying binomial theorem, and neglecting squares and higher powers, we get
$\frac{\text{T}'}{\text{T}}=1+\frac{1}{2}\times0.0005$
$\frac{\text{T}'}{\text{T}}-1=0.00025$
$\frac{\text{T}'-\text{T}}{\text{T}}=0.00025$
$\therefore$ Loss of time pre second $=0.00025\text{s}$
Loss of time per day
$=0.00025\times24\times60\times60\text{s}$
$=21.6\text{s}$
View full question & answer→Question 375 Marks
The length of a second’s pendulum on the surface of Earth is 1m. What will be the length of a second’s pendulum on the moon?
AnswerA pendulam of time period (T) of 2sec is called secound penduluam.
$\text{T}_\text{e}=2\pi\sqrt{\frac{\text{l}_\text{e}}{\text{g}_\text{e}}}\Rightarrow\text{T}_\text{e}^2=4\pi^2\frac{\text{l}_\text{e}}{\text{g}_\text{e}}\ ...(1)$
$\text{T}_\text{m}=2\pi\sqrt{\frac{\text{l}_\text{m}}{\text{g}_\text{m}}}\because\text{g}_\text{m}=\frac{\text{g}_\text{e}}{6}$
$\therefore\text{T}_\text{m}^2=4\pi^2\frac{\text{i}_\text{m}\times6}{\text{g}_\text{e}}\ ...(2)$
For secound pendulum $\text{T}_\text{e}=\text{T}_\text{m}=2\text{sec}$
$4\pi^26\text{l}_\text{m}$
$\frac{\text{T}_\text{m}^2}{\text{T}^2_\text{e}}=\frac{\text{g}_\text{e}}{4\pi^2\frac{\text{l}_\text{e}}{\text{g}_\text{e}}}\ \text{or}\ \frac{(2)^2}{(2)^2}=\frac{6\text{l}_\text{m}}{\text{l}_\text{e}}\text{l}_\text{e}=1\text{m}$
$\frac{1}{1}=\frac{6\text{l}_\text{m}}{1\text{m}}\Rightarrow\text{l}_\text{m}=\frac{1}{6}\text{m}$
View full question & answer→Question 385 Marks
A particle of mass 0.8kg is executing simple harmonic motion with amplitude of 1.0 metre and periodic time $\frac{11}{7}\text{sec}$. Calculate the velocity and the kinetic energy of the particle at the moment when its displacement is 0.6 metre.
AnswerWe know that, $\text{v}=\omega\sqrt{(\text{a}^2-\text{y}^2)}$ Further $\omega=\frac{2\pi}{\text{T}}$$\therefore\text{v}=\frac{2\pi}{\text{T}}\sqrt{(\text{a}^2-\text{y}^2)}$
$=\frac{2\times30.14}{\big(\frac{11}{7}\big)}\sqrt{[(1.0)^2-(0.6)^2]}$
$=3.2\text{m/ sec}$
Kinetic energy at displacement is given by$\text{K}=\frac{1}{2}\text{mv}^2=\frac{1}{2}\times0.8\times(3.2)^2$
$=4.1\text{ joule}$.
View full question & answer→Question 395 Marks
A simple harmonic motion has an amplitude A and time period T. What is the time taken to travel from $\text{x}=\text{A}$ $\text{x}=\frac{\text{A}}{2}$?
AnswerDisplacement from mean position.$=\text{A}-\frac{\text{A}}{2}=\frac{\text{A}}{2}$
Now $\text{y}=\text{A}\cos\omega\text{t}$$\frac{\text{A}}{2}=\text{A}\cos\frac{2\pi}{\text{T}}\text{t}$
$\cos\frac{2\pi}{\text{T}}\text{t}=\frac{1}{2}$
$\Rightarrow\cos\frac{2\pi}{\text{T}}\text{t}=\cos\frac{1}{3}$
$\frac{2\pi}{\text{T}}\text{t}=\cos\frac{1}{3}$
$\therefore\text{t}=\frac{\pi}{6}$
View full question & answer→Question 405 Marks
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
AnswerThe restoring force, F = -weight of mercury column of the height 2y ⇒ F = -(volume) × density × g$=-(\text{A}\times2\text{y}\rho\text{g})$
$\Rightarrow\text{F}=-2\text{A}\rho\text{gy} \ ...(\text{i})$
where A = Area ofcross-section ofthe tube, $\rho$ = densrty of mercury$\text{k}=-\frac{\text{F}}{\text{y}}=2\text{A}\rho.\text{g}$
Time period $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{\text{m}}{2\text{A}\rho\text{g}}}.$ Let L = length of the whole mercury column therefore, mass of mercury.$\text{m}=\text{volume}\times\text{density = AL}\rho$
$\text{T}=2\pi\sqrt{\frac{\text{AL}\rho}{2\text{A}\rho\text{g}}}$
$=2\pi\sqrt{\frac{\text{L}}{2\text{g}}}$
where L is the total length of mercury column of L = 2h. Where h is the height of mercury column in U-tube. It shows that mercury column executes S.H.M.
View full question & answer→Question 415 Marks
A body executing linear SHM has a velocity of $3cms^{-1}$ when its displacement is $4cm$ and a velocity of $4cms^{-1}$ when its displacement is $3cm$.
- Find the amplitude and period of the oscillation.
- If the mass of the body is $50g$, calculate the total energy of oscillation.
Answer
- In SHM, the velocity V at a displacement x is given by
$\text{V}=\omega(\text{A}^2-\text{x}^2)^\frac{1}{2}$
$\text{V}^2=\omega^2(\text{A}^2-\text{x}^2)$
$V = 3cms^{-1}$ when x = 4cm. Therefore
$9=\omega^2(\text{A}^2-16) \cdots\text{(i)}$
$V = 4cms^{-1}$ when x = 3cm. Therefore
$16=\omega^2(\text{A}^2-9) \cdots\text{(ii)}$
Simultaneous solution of eqns (i) and (ii) gives
Amplitude A = 5cm
Angular frequency, $\omega=1\text{ rad s}^{-1}$
Hence time period, $\text{T}=\frac{2\pi}{\omega}$
$=2\pi\text{ seconds}$
$=6.28\text{s}$
- $\text{m}=50\text{g}=50\times10^{-3}\text{kg}$
$\text{A}=5\text{cm}=5\times10^{-2}\text{m}$
$\omega=1\text{rad}\text{ s}^{-1}$
Total energy $=\frac{1}{2}\text{m}\text{A}^2\omega^2$
$=\frac{1}{2}\times(50\times10^{-3})\times5\times10^{-2})^2(1)^2$
$=6.25\times10^{-5}\text{J}$ View full question & answer→Question 425 Marks
Springs of spring constants $k, 2k, 4k, 8k$ ............ are connected in series. A mass m kg is attached to the lower end of the last spring and the system is allowed to vibrate. What is the time period of oscillations? Given $m = 40gm; k = 20Ncm^{-1}$
AnswerHere $m = 40g = 0.04kg; k = 2.0Ncm^{-1} 2.0 \times 100Nm^{-1}$ Effective spring constant K is given by$\frac{1}{\text{k}}=\frac{1}{\text{k}}+\frac{1}{2\text{k}}+\frac{1}{4\text{k}}+\frac{1}{8\text{k}}+.......$
$=\frac{1}{\text{k}}\big[1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.......\big]$
$=\frac{1}{\text{k}}\bigg[\frac{1}{1-\frac{2}{2}}\bigg]=\frac{2}{\text{k}}$
$=\text{k}=\frac{\text{k}}{2}$
$\therefore$ Time period $\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$
$\text{T}=2\pi\sqrt{\frac{\text{m}}{\frac{\text{k}}{2}}}$
$=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$.
$=2\times\frac{22}{7}\times\sqrt{\frac{2\times0.04}{2.0\times100}}$
$=0.126\text{s}$
View full question & answer→Question 435 Marks
A particle P starts its motion at t = 0 and moves along a circle as shown in figure with period of 4s. Write down the equation representing S.H.M. of the particle P.
Answer
At t = 0, the angle made by Op with x-axis is $+\pi$ radian, so $\phi=\pi$ Let P (t) be the position of the particle at time t. The projection of P (t) on XOX' diameter is M.$\therefore\text{x}=\text{op}\text{(t)}\cos(\omega\text{t}+\phi)$
$=4\cos\Big(\frac{2\pi}{\text{T}}\text{t}+\phi\Big)$
Since $\text{T}=4\text{s}$ $\phi=+\pi$
$\therefore\text{n}=4\cos\big(\frac{\pi}{2}\text{t}+\pi\big)$
$=-4\cos\big(\frac{\pi}{2}\text{t}\big)$
$(\because\cos(\pi+\theta)=-\cos\theta)$ View full question & answer→Question 445 Marks
A particle executes SHM of period 8s. After what time of its passing through the mean position, will be energy be half kinetic and half potential?
AnswerGiven $\text{PE}=\text{KE}$ i.e. $\frac{1}{2}\text{m}\omega^2\text{x}^2$$=\frac{1}{2}\text{m}\omega^2(\text{A}^2-\text{x}^2)$
$\text{x}^2=\text{A}^2-\text{x}^2$
$\Rightarrow\text{x}=\frac{\text{A}}{\sqrt{2}}$
Now, $\text{x}=\text{A}\sin\omega\text{t}$$=\text{A}\sin\Big(\frac{2\pi}{\text{T}}\Big)\text{t}$
So, $=\frac{\text{A}}{\sqrt{2}}$$=\text{A}\sin2\pi\frac{\text{t}}{8}$
$=\sin\frac{\pi\text{t}}{4}$
$=\frac{1}{\sqrt{2}}$
$=\sin\frac{\pi}{4}$
$=\frac{\pi\text{t}}{4}=\frac{\pi}{4}$
$\text{t}=1\text{s}$
View full question & answer→Question 455 Marks
The motion of a particle executing simple harmonic motion is described by the displacement function,$\text{x(t)}=\text{A}\cos(\omega\text{t}+\phi).$
If the initial (t = 0) position of the particle is 1cm and its initial velocity is $\omega\text{ cm/s,}$ what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi\text{s}^{-1}.$ If instead of the cosine function, we choose the sine function to describe the SHM: $\text{x}=\text{B}\sin(\omega\text{t}+\alpha),$ what are the amplitude and initial phase of the particle with the above initial conditions.
AnswerInitially, at t = 0: Displacement, x = 1cm Initial velocity, $\text{v}=\omega\text{ cm/sec.}$ Angular frequency, $\omega=\pi\text{ rad/s}^{-1}$ It is given that:$\text{x(t)}=\text{A}\cos(\omega\text{t}+\phi)$
$1=\text{A}\cos(\omega\times0\times\phi)$
$\text{A}\cos\phi=1\ ........(1)$
$\text{Velocity, }\upsilon=\frac{\text{dx}}{\text{dt}}$
$\omega=-\text{A}\omega\sin(\omega\text{t}+\phi)$
$1=-\text{A}\sin(\omega\times0+\phi)=-\text{A}\sin\phi$
$\text{A}\sin\phi=-1\ ....(2)$
Squaring and adding equations (1) and (2), we get:$\text{A}^2(\sin^2\phi+\cos^2\phi)=1+1$
$\text{A}^2=2$
$\therefore\ \text{A}=\sqrt{2}\ \text{cm}$
Dividing equation (2) by equation (1), we get:$\tan\phi=-1$
$\therefore\ \phi=\frac{3\pi}{4},\frac{7\pi}{4},....$
SHM is given as:$\text{x}=\text{B}\sin(\omega\text{t}+\alpha)$
Putting the given values in this equation, we get:
$1=\text{B}\sin[\omega\times0+\alpha]$
$\text{B}\sin\alpha=1\ .....(3)$
Velocity, $\upsilon=\omega\text{B}\cos(\omega\text{t}+\alpha)$ Substituting the given values, we get:$\pi=\pi\text{B}\sin\alpha$
$\text{B}\sin\alpha=1\ .....(4)$
Squaring and adding equations (3) and (4), we get:$\text{B}^2[\sin^2\alpha+\cos^2\alpha]=1+1$
$\text{B}^2=2$
$\therefore\ \text{B}=\sqrt{2}\text{ cm}$
Dividing equation (3) by equation (4), we get:$\frac{\text{B}\sin\alpha}{\text{B}\cos\alpha}=\frac{1}{1}$
$\tan\alpha=1=\frac{\tan\alpha}{4}$
$\therefore\ \alpha=\frac{\pi}{4},\frac{5\pi}{4}, .....$
View full question & answer→Question 465 Marks
The number of harmonic components in the oscillations are represented by, $\text{y}=4\cos^2$ at $\sin4\text{t}$. What are What are their corresponding angular frequencies?
Answer$\text{y}=4\cos^22\text{t}\sin4\text{t}$$=2(\cos4\text{t}+1)\sin4\text{t}$
$[\because2\cos^2\theta=\cos2\theta+1]$
$=2\sin4\text{t}\cos4\text{t}+2\sin4\text{t}$
$=\sin8\text{t}+2\sin4\text{t}$
$=2\sin4\text{t}+\sin8\text{t}$
The resulting harmonic oscillation is a combination of two harmonic motions of angular frequencies 4rad/ s and 8rad/ s.
View full question & answer→Question 475 Marks
A particle is executing SHM. If $v_1$ and $v_2$ are the speeds of the particle at distance $x_1$ and $x_2$ from the equilibrium position, show that the frequency of oscillations is $\text{f}=\frac{1}{2\pi}\Big(\frac{\text{v}_1^2-\text{v}_2^2}{\text{x}_2^2-\text{x}_1^2}\Big)^\frac{1}{2}$
AnswerThe displacement of a particle executing SHM is given by$\text{x}=\text{A}^2\cos\omega\text{t}$
$\frac{\text{dx}}{\text{dt}}=-\omega\text{A}\sin\omega\text{t}$
$\therefore$ Velocity $\text{v}=\frac{\text{dx}}{\text{dt}}$
$\text{v}^2=\text{A}^2\omega^2\sin^2\omega\text{t}$
$=\text{A}^2\omega^2(1-\cos^2\omega\text{t})$
$=\omega^2(\text{A}^2-\text{x}^2)$
View full question & answer→Question 485 Marks
A point particle of mass $0.1kg$ is executing S.H.M. of amplitude of $0.1m$. When the particle passes through the mean position, its kinetic energy is $8 \times 10^{-3}$ joule. Obtain the equation of motion of this particle if the initial phase of oscillation is $45^\circ$
AnswerThe displacement of a particle in S.H.M is given by$\text{y}=\text{a}\sin(\omega\text{t}+\phi)$
Velocity $=\frac{\text{dy}}{\text{dt}}=\omega\text{a}\cos(\omega\text{t}+\phi)$ The velocity is maximum when the particle passes through the mean position i.e.$=\Big(\frac{\text{dy}}{\text{dt}}\Big)_\text{max}=\omega\text{a}$
The kinetic energy at this instant is given by$=\frac{1}{2}\text{m}\Big(\frac{\text{dy}}{\text{dt}}\Big)^2$
$=\frac{1}{2}\text{m}\times\omega^2\text{a}^2$
$=8\times10^{-3}\text{ joule}$
$\frac{1}{2}\times(0.1)\omega^2\times(0.1)^2$
$=8\times10^{-3}\text{ joule}$
Solving we get $\omega=\pm4$ Substituting the values of a $\omega$ and $\phi$ in the equation of SHM., we get$\text{y}=0.1\sin(\pm4\text{t}+\frac{\pi}{4})\text{ metre}$
View full question & answer→Question 495 Marks
The displacement of two particles executing simple harmonic motion are represented by equations, $\text{y}=4\sin(10\text{t}+\theta)$ and $\text{y}=5\cos10\text{t}$ What is the phase difference between the velocities of these particles?
AnswerFor 1st particle$\text{y}_1=4\sin(10\text{t}+\theta);$
Velocity $\frac{\text{dy}_1}{\text{dt}}$$=4\times10\cos(10\text{t}+\theta)$
$=40\cos(10\text{t}+\theta)$ For second particle $\text{y}_2=5\cos10\text{t}=5\sin(10\text{t}+\frac{\pi}{2})$
Velocity $\frac{\text{dy}_2}{\text{dt}}$$=5\cos\times10\cos(10\text{t}+\frac{\pi}{2})$
$=5\cos(10\text{t}+\frac{\pi}{2})$
Phase difference between velocities $=(10\text{t}+\theta)-(10\text{t}+\frac{\pi}{2})$
$=\Big(\theta-\frac{\pi}{2}\Big)$
View full question & answer→Question 505 Marks
A body of mass $12kg$ is suspended by coil spring of natural length $50cm$ and force constant $2.0 \times 10^3Nm^{-1}$. What is the stretched length of the spring? If the body is pulled down further strething the spring to a length of $5.9cm$ and then released, then what is the frequency of oscillation of the suspended mass? (Neglect the mass of the spring)
AnswerGiven $m = 12kg$, Original length $l = 50cm, k = 20 \times 10^3Nm^{-1}$ As, F = ky$\therefore\text{y}=\frac{\text{F}}{\text{k}}$
$=\frac{\text{mg}}{\text{k}}$
$=\frac{12\times9.8}{2\times10^3}$
$=5.9\times10^{-2}\text{m}$
$=5.9\text{cm}$
$\therefore$ Stretched length of the spring $=\text{l}+\text{y}$
$=50+55.9\text{cm}$
$=105.9\text{cm}$
Frequency of oscillations, $\text{V}=\frac{1}{\text{T}}$$=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
$=\frac{1}{2\times3.14}\sqrt{\frac{2\times10^3}{12}}$
$=2.06\text{s}^{-1}$
View full question & answer→