3 Marks Question - Page 2 - Physics STD 11 Science Questions - Vidyadip

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3 Marks Question

Question 513 Marks

A uniform U-tube has a liquid of density $\rho$ and a length L. The cross-sectional area is A. If it is made to oscillate, show that it will be S.H.M. and find its frequency.

Answer

Let L be the length of the liquid column in the U-tube of uniform cross section A. The liquid density be $\rho.$ If by tilting, the level of liquid in the limbs differ by 2x, the excess pressure on the higher level side from the same height as the other equalling $2\text{x}\rho\text{g}$ provides restoring force. The entire liquid oscillates as a result.

Since, restoring force = mass × acceleration.

$=-2\text{x}\rho\text{gA = AL}\rho\text{a,a}=-\frac{2\text{gx}}{\text{L}}$

The frequency $=\text{f}=\frac{1}{2\pi}\sqrt{\frac{\text{acceleration}}{\text{displacement}}}$

$=\frac{1}{2\pi}\sqrt{\frac{2\text{g}}{\text{L}}}$

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Question 523 Marks

A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 2.5cm. What must be the least period of these oscillations so that the object is not detached from the platform? Take g = 10ms^-2?

Answer

The object will not detach from the platform, if the angular frequency $\omega$is such that, during the downward motion, the maximum acceleration equals the acceleration due to gravity, i.e.,

$\omega^2_\text{max}\text{A}=\text{g}$

$\omega_\text{max}=\sqrt{\frac{\text{g}}{\text{A}}}$

$\text{T}_\text{min}=\frac{2\pi}{\omega_\text{max}}$

$=2\pi\sqrt{\frac{\text{A}}{\text{g}}}$

Now $\text{A}=2.5\text{cm}$

$=2.5\times10^{-2}\text{m}$

$\text{g}=10\text{ms}^{-2}$

Substituting thesa values we get

$\text{T}_\text{min}=\frac{\pi}{10}$

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Question 533 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?

Answer

Equivalent spring constant: k' = 2k

= 1200Nm^-1, m = 3kg

$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{3}{120}}$

$=\frac{2\pi}{20}=\frac{\pi}{10}\text{sec}.$

Maximum speed

$\text{v}=\omega\text{A}=20\times5\times10^{-2}=1\text{ms}^{-1}$

Energy dissipated = Maximum energy

$=\frac{1}{2}\text{m}\omega^2\text{A}^2$

$=\frac{1}{2}\times3\times20^2\times25\times10^{-4}$

$=600\times25\times10^{-4}$

$=150\times10^{-2}\text{Joule}$

$=1.5\text{ Jule}$

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Question 543 Marks

A particle of mass 10g is placed in a potential field given by U = 50x² + 100 erg/gm. Calculate the frequency of oscillation.

Answer

P.E. of 10 gram particle is U

= 10(50x² + 100) erg. The force acting on the particle is given by

$\text{F}=\frac{-\text{dU}}{\text{dx}}=\frac{-\text{d}}{\text{dx}}(500\text{x}^2+1000)$

$=-1000\text{x}$

But $\text{F = m}\frac{\text{d}^2\text{x}}{\text{dt}^2}$

$\therefore\text{m}\frac{\text{d}^2\text{x}}{\text{dt}^2}=-1000\text{x}$

$\frac{\text{d}^2\text{x}}{\text{dt}^2}=\frac{-1000\text{x}}{\text{m}}=\frac{-1000\text{x}}{10}$

$=-100\text{x}$

As $\frac{\text{d}^2\text{x}}{\text{dt}^2}=\omega^2\text{x};$

so $\omega^2=100$

$\omega=\sqrt{100}=10$

Frequency of oscillation,

$\text{v}=\frac{\omega}{2\pi}=\frac{\sqrt{100}}{2\pi}=1.58\text{s}^{-1}$

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Question 553 Marks

The bottom of a dip on a road has a radius of curvature R. A rickshaw of mass M left a little away from the bottom oscillates about this dip. Deduce an expression for the period of oscillation.

Answer

Let R be the radius of the dip, and O be its centre. Let the rickshaw of mass M be at P at any instant. This case is similar to that of a simple pendulum. The force that produces oscillations in the rickshaw is $\text{F}=\text{Mg}\sin\theta.$ If $\theta$ is small and is measured in radian then, $\sin\theta=\theta,$

$\therefore\text{F}=-\text{Mg}\theta$ ($\because$ force acts to reduce $\theta$)

Displacement of the rickshaw OP $=\text{y}=\text{R}\theta$

$\therefore$ Force constant,

$\text{k}=\frac{-\text{Force}}{\text{Displacement}}$

$=-\Big(\frac{-\text{Mg}\theta}{\text{R}\theta}\Big)=\frac{\text{Mg}}{\text{R}}$

$\therefore$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{Inertial factor}}{\text{Spring factor}}}$

$=2\pi\sqrt{\frac{\text{MR}}{\text{Mg}}}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}$

$\therefore\text{T}=2\pi\sqrt{\frac{\text{R}}{\text{g}}}$

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Question 563 Marks

A body of mass 1.0kg is suspended from a weightless spring having force constant 600Nm^-1. Another body of mass 0.5kg moving vertically upwards hits the suspended body with a velocity of 3.0ms^-1 and gets embedded in it. Find the frequency of oscillations and amplitude of motion.

Answer

Here, inertia factor = Oscillating mass = (m + m₁)

= 1 + 0.5 = 1.5kg

Spring factor = Force conslant

= K = 600Nm^-1

Frequency of oscillation,

$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{Spring factor}}{\text{Inertial factor}}}$

$=\frac{10}{\pi}\text{Hz}$

Let v₁ be the velocity ofthe mass (m + m₁) after collision.

According to law of conservation of linear momentum, we have

$\text{m}_1\text{v}=(\text{m + m}_1)\text{v}_1$

$\text{v}_1=\frac{\text{m}_1\text{v}}{\text{m + m}_1}$

$\frac{\text{m}_1\text{v}}{\text{m + m}_1}=\frac{0.5\times3}{1+0.5}=1\text{ms}^{-1}$

Here the collision is inelastic. According to the law of conservation of mechanical energy, we have

$(\text{K.E}.)_{\text{max}}=(\text{P.E.})_{\text{max}}$

i.e., $\frac{1}{2}(\text{m + m}_1)\text{v}^2_1=\frac{1}{2}\text{kA}^2$

$\text{A}=\text{v}_1\sqrt{\frac{\text{m + m}_1}{\text{k}}}$

$=\sqrt{\frac{1.5}{600}}=\frac{1}{20}=5\text{cm}$

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Question 573 Marks

Prove that if a liquid taken in a U-tube is disturbed from the state of equilibrium, it will oscillate simple harmonically. Find expressions for the angular frequency and time period.

Answer

The restoring force,

F = weight of liquid column of the height 2y

$\Rightarrow\text{F}=-(\text{volume})\times\text{density}\times\text{g}$

$=-(\text{A}\times-2\text{y}.\rho\text{g})$

$\Rightarrow\text{F}=-2\text{A}\rho\text{g.y} \ ...(\text{i})$

where A = Area of cross-section of the tube

$\rho$ = density 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

$\text{m}=\text{volume}\times\text{density}=\text{A.L.}\rho.$

$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{\text{A.L.}\rho}{2\text{A}\rho\text{g}}}=2\pi\sqrt{\frac{\text{L}}{2\text{g}}}$ where L is the totat 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.

Frequency, $\text{v}=\frac{1}{\text{T}}=\frac{1}{2\pi}\sqrt{\frac{2\text{g}}{\text{L}}}$

So, angular frequency

$\omega=2\pi\text{v}=2\pi\times\frac{1}{2\pi}\sqrt{\frac{2\text{g}}{\text{L}}}=\sqrt{\frac{2\text{g}}{\text{L}}}$

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Question 583 Marks

Aman stands on a weighing machine placed on a horizontal platform. The machine reads 50kg. By means of a suitable mechanism, the platform is made to execute harmonic vibrations up and down with a frequency of two vibrations per second. What will be the effect on the reading of the weighing machine? The amplitude of vibrations of platform is 5cm. Take, g = 10ms^-2.

Answer

Here. m = 50kg. v = 2s^-1

A = 5cm = 0.05m

Maximum acceleration

$\text{a}_\text{max}=\omega^2\text{A}$

$=(2\pi\text{v}^2)\text{A}=4\pi^2\text{v}^2\text{A}$

$=4\times\Big(\frac{22}{7}\Big)^2\times(2)^2\times0.05$

$=7.9\text{cm}^{-2}$

$\therefore$ Maximum force felt by the man $=\text{m}(\text{g}+\text{a}_\text{max})$

$=50(10+7.9)$

$=895.0\text{N}$

$=89.5\text{kgf}$

Minimum force felt by the man $=\text{m}(\text{g}-\text{a}_\text{max})$

$=50(10-7.9)$

$=105.0\text{N}$

$=10.5\text{kgf}$

Hence, the reading of the weighing machine varies between 10.5kgf and 69.5kgf.

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Question 593 Marks

At a time when the displacement is half the amplitude, what fraction of the total energy is kinetic and what fraction is the potential in S.H.M?

Answer

When $\text{x}=\frac{\text{A}}{2},$

Fraction of energy as potential is

$=\frac{\frac{1}{2}\text{m}\omega^2\Big(\text{A}^2-\frac{\text{A}^2}{4}\Big)}{\frac{1}{2}\text{m}\omega^2\text{A}^2}=\frac{3}{4}\text{ or }75\%$

So, fraction of energy as kinetic is

$=\frac{\frac{1}{2}\text{m}\omega^2\frac{\text{A}^2}{4}}{\frac{1}{2}\text{m}\omega^2\text{A}^2}=\frac{1}{2}\text{ or }25\%$

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Question 603 Marks

A spring of force constant k has a mass M suspended from it. If the spring is cut into two halves, and the same mass is attached to one of the pieces, what will be the frequencies of oscillation of the mass?

Answer

When the spring is cut into two equal halves, the force constant of each part will be doubled. Therefore, the original frequency, $\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{M}}}$ will become $\text{v}'=\frac{1}{2\pi}\sqrt{\frac{2\text{k}}{\text{M}}}=\sqrt{2\text{v}}$

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Question 613 Marks

Calculate the percentage change in time period of a simple pendulum if its length is increased by 8%.

Answer

As $\text{T}\propto\sqrt{\text{l}}$

$\frac{\Delta\text{T}}{\text{T}}\times100=\frac{1}2{}\frac{\Delta\text{l}}{\text{l}}\times100$

$\therefore\%\text{ change in T}=\frac{1}{2}\times8=4\%$

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Question 623 Marks

A particle of mass 0.1kg is held between two rigid supports by two springs of force constants 8N/ m and 2N/ m. If the particle is displaced along the direction of the length of the springs, calculate its frequency of vibration.

Answer

The situation is shown in the fig.

When the mass is displaced along the direction of the length of the spring, one spring is compressed while the other is extended but the force due to both the springs is in the same direction. Hence the effective force constant

k = k₁ + k₂ = 8N/ m + 2N/ m = 10N/ m

The frequency of vibration is given by

$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$

$=\frac{1}{2\pi}\sqrt{\frac{10}{0.1}}$

$\text{v}=\frac{10}{2\pi}$

$=\frac{5}{\pi}\text{s}^{-1}$

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Question 633 Marks

Show that when a particle is moving in S.H.M. its velocity at a distance $\frac{\sqrt{3}}{2}$ of its amplitude from the central position is half its velocity in central position.

Answer

In a S.H.M. velocity is given by $\text{v}=\omega\sqrt{\text{A}^2-\text{x}^2}$ where x is the displacement from mean position.

Velocity at $\text{x}=\frac{\sqrt{3}\text{A}}{2}$ is,

$\text{v}_1=\omega\sqrt{\text{A}^2-\frac{3}{4}\text{A}^2}=\omega\text{A}\sqrt{\frac{1}{4}}$

$=\frac{\omega\text{A}}{2}$

Velocity at central position

$=\omega\sqrt{\text{A}^2-0^2}=\omega\text{A}$

$\therefore$ Velocity at $\frac{\sqrt{3}\text{A}}{2}=\frac{1}{2}$ (velocity at central position)

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Question 643 Marks

Explain the relation in phase between displacement, velocity and acceleration in S.H.M., graphically as well as theoretically.

Answer

If displacement:

$\text{y = A}\sin(\omega\text{t}+\phi_0)$

velocity $\text{v}=\omega\text{A}\cos(\omega\text{t}+\phi_0)$

acceleration $\text{a}=-\omega^2\text{A}\sin(\omega\text{t}+\phi_0)$

x and v differ in phase by $\frac{\pi}{2}.$

v and a differ in phase by $\frac{\pi}{2}.$

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Question 653 Marks

When will the motion of a simple pendulum be simple harmonic?

Answer

Consider a pendulum of length l and mass of bob m is displaced by angle $\theta$ as shown in fig.

The restoring force = F $=\text{mg}\sin\theta$

If $\theta$ is small then $\sin\theta=\theta=\frac{\text{arc}}{\text{radius}}=\frac{\text{x}}{\text{l}}$

$\therefore\text{F}=-\text{mg}\frac{\text{x}}{1}\ \text{or}\ \text{F}\propto(-\text{x})$

$(\because\text{m}_2\text{g},1\text{are constant})$

Hence, the motion of simple pendulam will be smile harmonic for small angle $\theta.$

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Question 663 Marks

Figure below shows four different spring arrangements. If the mass in each arrangement is displaced from its equilibrium position and released, what is the resulting frequency of vibration in each case? Neglect the mass of the spring. (Fig. (a) and (b) represent an arrangement of springs in parallel, and (c) and (d) represent springs in series)

Answer

(a), (b) In parallel equivalent value of k = k₁ + k₂

$\therefore\text{f}=\frac{1}2\pi{}\sqrt{\frac{\text{k}_1+\text{k}_2}{\text{m}}}$

(c), (d) In series equivalent $\text{k}=\frac{\text{k}_1\text{k}_2}{\text{k}_1+\text{k}_2}$

$\therefore\text{f}=\frac{1}{2\pi}\sqrt{\frac{\text{k}_1\text{k}_2}{(\text{k}_1+\text{k}_2)\text{m}}}$

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Question 673 Marks

A 0.2kg. of mass hangs at the end of a spring. When 0.02kg more mass is added to the end of the spring, it stretches 7cm more. If the 0.02kg mass is removed, what will be the period of vibration of the system?

Answer

When 0.02kg is added, there is a stretch of 7cm. Using

mg = Kx, we have

$\text{K}=\frac{0.02\times10}{7\times10^{-2}}=\frac{20}{7}=2.86\text{N/m}$

Time period $=\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{K}}}$

$=2\pi\sqrt{\frac{0.2}{2.86}}=1.66\sec.$

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Question 683 Marks

A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?

Answer

The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

Acceleration due to gravity = g

Centripetal acceleration = v²/R

where,

v is the uniform speed of the car

R is the radius of the track

Effective acceleration (g') is given as:

$\text{g}'=\sqrt{\text{g}^2+\frac{\upsilon^4}{\text{R}^2}}$

$\therefore$ Time period, $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}'}}$

$=2\pi\frac{\text{l}}{\text{g}^2+\frac{\upsilon^4}{\text{R}^2}}$

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Question 693 Marks

A spring compressed by 0.1m develops a restoring force of 10N. A body of mass 4kg is placed on it. Deduce the (i) force constant of the spring (ii) depression of the spring under the weight of the body and (iii) period of oscillation, if the body is disturbed.

Answer

Restoring force,

F = 10N;

Mass of body $\text{m = 4kg}$

Displacement $\xi=0.1\text{m}$

The force constant of spring

$\text{k}=\frac{\text{Force}}{\text{Displacement}}=\frac{10}{0.1}$

$=100\text{Nm}^{-1}$

Depression due lo weight

$=\frac{\text{Force}}{\text{k}}=\frac{4\times10}{100}=0.4\text{m}$

Period of oscillation,

$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}=2\pi\sqrt{\frac{4}{100}}=\frac{2\pi}{5}\text{s}$

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Question 703 Marks

Derive an expression for the potential energy of an elastic stretched spring.

Answer

Consider a spring attached with a mass m stretching by a length y after t sec.

Restoring force

F = mass × acceleration

$=-\text{}\omega^2\text{y}=-\text{ky}$

where, $\text{k = spring constant = m}\omega^2$

Work done for an additional displacement dy against restoring force is

$\text{dW}=-\text{F dy}$

$=-(-\text{ky})\text{dy = ky dy}$

Total work done

$\text{W}=\int\limits^{\text{y}}_0\text{ky dy}=\frac{1}{2}\text{ky}^2$

This work done appears as a PE. 'U' of the particle.

$\text{U}=\frac{1}2{}\text{ky}^2=\frac{1}{2}\text{m}\omega^2\text{y}^2$

$=\frac{1}{2}\text{m}\omega^2\text{a}^2\sin^2\omega\text{t}$

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Question 713 Marks

What is a second's pendulum? How much is its length on the surface of moon?

Answer

A second's pendulum is one whose time period of e. oscillation is 2 seconds. On the surface of moon,

$\text{a}=\frac{\text{g}}{6}.$

$\therefore\text{Using}\text{ T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}},$

We have $\text{l}=\frac{4\text{g}}{6\times4\pi^2}=\frac{1}{6}\text{m}$

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Question 723 Marks

Show that for a particle in linear SHM, the average kinetic energy over a period of oscillation is equal to average potential energy over the same period.

Answer

K.E. $=\frac{1}{2}\text{m}\omega^2\text{A}^2\cos^2\omega\text{t}$

P.E. $=\frac{1}{2}\text{m}\omega^2\text{A}^2\sin^2\omega\text{t}$

Average K.E. over a one time period:

$(\text{K.E.)}_{\text{avg}}=\frac{1}{\text{T}}\int\limits^{\text{T}}_0(\text{K.E.)dt}$

$=\frac{1}{\text{T}}\int\limits^{\text{T}}_{0}\frac{1}{2}\text{m}\omega^2\text{A}^2\cos^2\omega\text{t dt}$

$=\frac{1}{2\text{T}}\text{m}\omega^2\text{A}^2\int\limits^\text{T}_0\cos^2\omega\text{t dt}$

$=\frac{\text{m}\omega^2\text{A}^2}{2\text{T}}.\frac{\text{T}}{2}=\frac{1}{4}\text{m}\omega^2\text{A}^2$

Similarly, $\text{P.E.}=\frac{1}{2}\text{m}\omega^2\text{A}^2$

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Question 733 Marks

Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).

$\text{x}=-2\sin\Big(3\text{t}+\frac{\pi}{3}\Big)$

Answer

$\text{x}=-2\sin\Big(3\text{t}+\frac{\pi}{3}\Big)=+2\cos\Big(3\text{t}+\frac{\pi}{3}+\frac{\pi}{2}\Big)$

$=2\cos\Big(3\text{t}+\frac{5\pi}{6}\Big)$

If this equation is compared with the standard SHM equation $\text{x}=\text{A}\cos\bigg(\Big(\frac{2\pi}{\text{T}}\Big)\text{t}+\phi\bigg),$ then we get:

Amplitude, A = 2cm

Phase angle, $\phi=\frac{5\pi}{6}=150^\circ$

Angular velocity, $\omega=\frac{2\pi}{\text{T}}=3\text{rad/sec.}$

The motion of the particle can be plotted as shown in the following figure.

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Question 743 Marks

A body oscillates with S.H.M. according to the equation:

$\text{x(t)}=5\cos\Big(2\pi\text{t}+\frac{\pi}{4}\Big)$

where x is in meters and t is in seconds.

Calculate the following:

Displacement at t = 0
Angular frequency
Magnitude of velocity (Maximum).

Answer

$\text{x(t)}=5\cos\Big(2\pi\text{t}+\frac{\pi}{4}\Big)\text{m}$

$\text{x}(0)=5\cos\Big(0+\frac{\pi}{4}\Big)=5\times\frac{1}{\sqrt{2}}\text{m}=\frac{5}{\sqrt{2}}\text{m}$
Angular frequency $=\omega=2\pi\text{v}\text{ rad/sec}.$
Maximum velocity $=\omega\text{A}=2\pi\times5=10\pi\text{ ms}^{-1}$

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Question 753 Marks

The potential energy of a particle of mass 1kg in motion along the x-axis is given by $\text{U}=4(1-\cos2\text{x})$ Here x is in metres. Find the period of small oscillations.

Answer

$\text{F}=-\frac{\text{dU}}{\text{dx}}$

$=-\frac{\text{d}}{\text{dx}}[4(1-\cos2\text{x})]-8\sin2\text{x}$

$\text{F}=-8\times2\text{x}$

[When x is small, $\sin2\text{x}=2\text{x}$]

$\text{F}=-16\text{x}$

As $\text{F}\propto\text{x}$ and -ve sign shows that x is directed towards equilibrium position hence the particle will execute SHM.

Here spring factor, k = 16N/ m

Inertia factor m = 1kg

$\therefore$ Time period $\text{T}=32\pi\sqrt{\frac{\text{m}}{\text{k}}}$

$=2\pi\sqrt{\frac{1}{16}}=\frac{\pi}{2}\text{s}$

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Question 763 Marks

A pendulum of length l is attached with a bob and placed in a lift. What will be time period when the lift is:

Having uniform motion upwards.
Accelerated upwards by a.
Accelerated downward by a?

Answer

When there is uniform motion there is no change in acceleration.

$\therefore\text{T}=\pi\sqrt{\frac{\text{l}}{\text{g}}}$

When there is an upward acceleration, g → g + a.

$\therefore\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g + a}}}$

When there is a downward acceleration g → g - a.

$\therefore\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}-\text{a}}}$

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Question 773 Marks

A simple pendulum in a stationary lift has time period T. What would be the effect on the time period when the left

Moves up with uniform velocity v.
Moves down with uniform velocity v.
Moves up with uniform acceleration a.
Moves down with uniform acceleration a.
Beings to fall freely under gravity?

Answer

And
Since acceleration of the lift is zero therefore there will be no effect on time period.
When the lift moves up with uniform acceleration a, the effective value of acceleration due to gravity is g + a

$\therefore\text{T}'=2\pi\sqrt{\frac{\text{l}}{\text{g}+\text{a}}}$

Clearly, $\text{T}'<\text{T}$.

When the lift moves down with uniform acceleration a, then the effective value of g is g - a.

$\text{T}'=2\pi\sqrt{\frac{\text{l}}{\text{g}+\text{a}}}$

Clearly,$\text{T}'<\text{T}$.

When the lift begins to fall freely under gravity, the effective value of g becomes zero. So, T is infinite i.e., the simple pendulum shall not oscillate.

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3 Marks Question - Page 2 - Physics STD 11 Science Questions - Vidyadip