Physics M.C.Q (1 Marks)

1. Simple harmonic motion.

3. Periodic motion.

Explanation:

For small angular displacement, the situation is shown in the figure. Only one restoring force creates.

motion in ball inside bowl.

Only one restoring force creates motion in ball inside bowl.

$\text{F}=-\text{mg}\sin\theta$

As $\theta\ \text{is small},\sin\theta=\theta$

So, $\text{ma}=-\text{mg}\frac{\text{x}}{\text{R}}$

Or, $\text{a}=-\Big(\frac{\text{g}}{\text{R}}\Big)\text{x}\Rightarrow\text{a}\propto-\text{x}$

So, motion of the ball is S.H.M and periodic.

1. $4\pi\times10^{-1}\text{s}$

Exlapation:

Here,m = 400g

400 × 10^-3kg

As R = 10Nm,^-1

^{$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{R}}}$}

^{$=2\pi\sqrt{\frac{400\times10^{-3}}{10}}$}

^{$=4\pi\times10^{-1}\text{s}$}

1. $2\pi$

Explanation:

A periodic function repeats itself after a time period T. and f(t) = f(t + T) As $\sin(\omega\text{t}+2\omega\pi)$ 

$\therefore$ Period of function is

1. $\pi\text{s}.$

Explanation:

Key concept: Let equation of an SHM is represented by $\text{y}=\text{a}\sin\omega\text{t}$

$\text{v}=\frac{\text{dy}}{\text{dt}}=\text{a}\omega\cos\omega\text{t}$

Hence $(\text{v})_{\text{max}}=\text{a}\omega$

Acceleration $(\text{A})=\frac{\text{dx}^2}{\text{dt}^2}=-\text{a}\omega^2\sin\omega\text{t}$

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

Maximum speed, $\text{v}_\text{max}=\omega\text{A}$

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

Divide eqn. (ii) by eqn. (i), we get

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

$\therefore\frac{\text{a}_\text{max}}{\text{v}_\text{max}}=\frac{2\pi}{\text{T}}$

Here, $\text{v}_\text{max}=30\text{cms}^{-1},\text{a}_\text{max}=60\text{cms}^2$

$\therefore\text{T}=2\pi\Big(\frac{30\text{cms}^{-1}}{60\text{cms}^{-2}}\Big)=\pi\text{s}$

1. 4.9cm

Explanation:

In equilibrium kx = mg

$\therefore$ Extension, $\text{x}=\frac{\text{mg}}{\text{k}}$

$\text{x}=\frac{20\times9.8}{4000}$

$\text{x}=0.049\text{m}$

$\text{x}=4.9\text{cm}$

1. Less than T

Explanation:

$\text{g}'=\text{g}+\frac{\text{F}_{\text{e}}}{\text{m}},\text{g}'>\text{g}$

Since, $\text{T}=\sqrt{\frac{\text{l}}{\text{g}}}$

So $\text{T}\propto\frac{1}{\sqrt{\text{g}}}$ and $\text{T}'\propto\frac{1}{\sqrt{\text{g}'}}$

$\therefore\text{T}'<\text{T}$

1. 10mm

Explanation:

$\text{v}_{\text{max}}=\text{r}\omega=\frac{\text{r}2\pi}{\text{T}}$

$\text{r}=\frac{\text{v}_{\text{max}}\text{T}}{2\pi}=\frac{10^3\times10^-5}{{2\times}\Big(\frac{22}{7}\Big)}=1.59\times10^{-3}\text{m}$

2. Simple harmonic with period $\frac{\pi}{\omega}.$

Explanation:

A simple harmonic motion is produced when a force (called restoring force) proportional to the displacement acts on a particle.

All sine and cosine functions of t are simple harmonic in nature.

Hence the motion is simple harmonic motion.

A simple harmonic motion is always periodic.

the motion is simple harmonic with time period $\frac{\pi}{\omega}.$

2. $\sqrt{\frac{10}{9\text{T}}}$

3. $\frac{2\text{mg}}{\text{k}}$

1. $\text{A = x}_0\omega^2,\delta=\frac{3\pi}{4}$

Explanation:

Acceleration, $\text{a}=\frac{\text{dv}}{\text{dt}}$

$=\text{x}_0\omega^2\cos\Big[\omega\text{t}+\frac{3\pi}{4}\Big]$

4. $\frac{1}{2}$

Explanation:

Here, a = 4cm; T = 12s. If t is the time takeo by particle in going fiom mean position to 2cm, then using

$\text{y}=\text{a}\sin\omega\text{t},$ we have $2=4\sin\frac{2\pi}{\text{T}}\text{t}$

$\sin\frac{2\pi}{\text{T}}\text{t}=\frac{2}{4}=\frac{1}{2}=\sin\frac{\pi}{6}$

$\text{t}=1\text{sec}.$

Time taken by particle to go from mean position to extreme position $=\frac{\text{T}}{4}=\frac{12}{4}=3\text{s}$

Therefore, time taken by Particle in going from 2cm to 4cm (i.e., extreme position)

$\text{t}'=3-1=2\text{s}$

$\therefore$ So $\frac{\text{t}}{\text{t}'}=\frac{1}{2}$

4. 2f

Exlanation:

Frequency of oscillation,

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

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

1. 0.62cm-s^-1

Explanation:

Equating the given equation with the equation of S.H.M.; $\text{y = r}\sin(\omega\text{t}+\theta)$

we have r = 1cm and $\theta=\frac{\pi}{4}$

$\therefore\text{Max. acceleration}=\omega^2\text{r}=\Big(\frac{\pi}{4}\Big)^2\times1=0.62\text{cm-s}^{-1}$ 

1. $\frac{6}{5}$

Explanation:

Given, $\text{y = kt}^2;\frac{\text{dy}}{\text{dt}}=2\text{kt}$ and

$\frac{\text{d}^2\text{y}}{\text{dt}^2}=2\text{k}=2\times1=2\text{ms}^{-2}$

So point of suspension is moving upwards with acceleration a = 2ms^-2. The effective acceleration due to gravity on pendulum is

$\text{g}'=\text{g + a}=10+2=12\text{ms}^{-2}$

$\text{T}_1=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ and $\text{T}_2=2\pi\sqrt{\frac{\text{l}}{\text{g}'}}$

$\therefore\frac{\text{T}^2_1}{\text{T}^2_2}=\frac{\text{g}'}{\text{g}}=\frac{12}{10}=\frac{6}{5}$

3. 3kg

Explanation:

$\text{v}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$ and $\frac{\text{v}}{2}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}_1}}$

$\therefore2=\sqrt{\frac{\text{m}_1}{\text{m}_2}}$

$\text{m}_1=4\text{m}=4\times1=4\text{kg}$

Hence mass added $=\text{m}_1-\text{m}=4-1=3\text{kg}$

2. k, A

Explanation:

Total energy of a particle performing S.H.M. is

$\text{E}=\frac{1}{2}\text{kA}^2$​​​​​​​

2. C will vibrate with maximum amplitude.

Explanation:

Here A is given a transverse displacement. Through the elastic support the disturbance is transferred to all the pendulums.

A and C are having same length, hence they will be in resonance, because of their time period of oscillation. Since length of pendulums A and C is same and $\text{T}=2\pi\sqrt{\frac{\text{L}}{\text{g}}}$ hence their time period is same and they will have frequency of vibration. Due to it, a resonance will take place and the pendulum C will vibrate with maximum amplitude.

3. $4\sqrt{2}$

Exlanation:

Given equation $\text{x(t)}=4(\cos\pi\text{t}+\sin\pi\text{t})$

Now comparing cbove equation with general form $\text{x(t)}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}$

We get A = 4 and B = 4

As, the resultant amplitude for such a equation is

$=\sqrt{\text{A}^2+\text{B}^2}$

$\therefore$ Amplitude $=\sqrt{4^2+4^2}=4\sqrt{2}$

1. Periodic motion.

3. Periodic but not simple harmonic motion.

Explanation:

Rotation of earth about its axis repeats its motion after a fixed interval of lime, so its motion is periodic.

The rotation of earth is obviously not a to and fro type of motion about a fixed point, hence its motion is not an oscillation. Also this motion does not follow S.H.M equation, $\text{a}\propto-\text{x}$

Hence, this motion is not a S.H.M.

2. $\sqrt{\frac{\text{k}}{\text{m}_2}}$

Explanation:

When mass m₁, is removd there is only mass m₂ attached to the spring of spring constant k. Then spring factor = k and inertia factor = m₂.

Angular frequency, $\omega=\sqrt{\frac{\text{Spring factor}}{\text{Inertia factor}}}=\sqrt{\frac{\text{k}}{\text{m}_2}}$

2. 2

Explanation:

In case of simple pendulum,

$\text{T}=2\pi\sqrt{\frac{5}{\text{g}}}$ and $\text{T}'=2\pi\sqrt{\frac{10}{\text{g}}}$

$\therefore\text{T}'=2\text{T}$

Let the two pendulums are in same phase for first time when shorter one has completed n oscillations. Then

$\text{nT = (n}-1)\text{T}'$

$\frac{\text{n}}{\text{n}-1}=\frac{\text{T}'}{\text{T}}=\frac{2\text{T}'}{\text{T}}=2$

$\text{n = 2n}-2$

$\text{n}=2$

1. 0.1256 sec.

Explanation:

The object is not detached from the platform if

$\text{mg = mr}\omega^2=\text{mr}\frac{4\pi\text{r}^2}{\text{T}^2}$

$\text{T}=2\pi\sqrt{\frac{\text{r}}{\text{g}}}=2\times\frac{22}{7}\sqrt{\frac{3.29\times10^{-3}}{9.8}}$

$=0.1256\text{ sec}$

2. 2cm s^-1

Explanation:

Here; f = 6s; r : 4cm; Displacement from the mean position when the panicle travels 2cm from the extreme position is, x = 4 - 2 = 2.0cm. If t is the time taken then,

$\text{x = r}\cos\omega\text{t = r}\cos\frac{2\pi}{\text{T}}\text{t}$

$\therefore2=4\cos\frac{2\pi}{6}\text{t}$

$\cos\frac{\pi\text{t}}{3}=\frac{2}{4}=\frac{1}{2}=\cos\frac{\pi}{3}$

$\text{t}=1\text{s}$

$\therefore$ Average velocity $=\frac{\text{Distance}}{\text{Time}}=\frac{2}{\text{T}}=2\text{cm s}^{-1}$

3. $\pi\sqrt{\frac{3\text{m}}{\text{k}}}$

1. $\text{Zero}$

1. Average total energy per cycle is equal to its maximum kinetic energy.

2. Average kinetic energy per cycle is equal to half of its maximum kinetic energy.

4. Root mean square velocity is $\frac{1}{\sqrt{2}}$ times of its maximum velocity.

Explanation:

In case of S.H.M, average total energy per cycle

= Maximum kinetic energy (K₀)

= Maximum potential energy (U₀)

Average KE per cycle $=\frac{0+\text{K}_0}{2}=\frac{\text{K}_0}{2}$

Let us write the equation for the SHM $\text{x}=\text{a}\sin\omega\text{t}.$

Clearly, it is a periodic motion as it involves sine function.

Let us find velocity of the particle, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\omega\text{t})=\text{a}\omega\cos\omega\text{t}$

Mean velocity over a complete cycle,

$\text{v}_\text{mean}=\frac{\int_{0}^{2\pi}\omega\text{a}\cos\theta\text{d}\theta}{2\pi}=\frac{\omega\text{a}[\sin\theta]^2_0}{2\pi}=0$

So, $\text{v}_\text{mean}\neq\frac{2}{\pi}\text{v}_\text{max}$

Root mean square speed,

$\text{v}_\text{ms}=\sqrt{\frac{\text{v}^2_\text{min}+\text{v}^2_\text{max}}{2}}=\sqrt{\frac{0+\text{v}^2-\text{max}}{2}}$

$\text{v}_\text{ms}=\frac{1}{\sqrt{2}}\text{v}_\text{max}$

4. The motion is S.H.M. with amplitude $\sqrt{\text{a}^2+\text{b}^2}.$

Explanation:

key concept: The sum of two S.H.Ms of same frequencies is a S.H.M.

According to the question, the displacement

$\text{y}=\text{a}\sin\omega\text{t}+\text{b}\cos\omega\text{t}$

Given $\text{x}=\text{a}\sin\omega\text{t}+\text{b}\cos\omega\text{t}$

Let $\text{a}=\text{A}\cos\phi$

And $\text{b}=\text{A}\sin\phi$

Squaring and adding (ii) and (iii), we get

$\text{a}^2+\text{b}^2=\text{A}^2\cos^2\phi+\text{A}^2\sin^2\phi=\text{A}^2$

$=\text{A}^2\Rightarrow\text{A}=\sqrt{\text{a}^2+\text{b}^2}$

$\text{y}=\text{A}\sin\phi.\sin\omega\text{t}+\text{A}\cos\phi.\cos\omega\text{t}$

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

$\frac{\text{dy}}{\text{dt}}=\text{A}\omega\cos(\omega\text{t}+\phi)$

$\frac{\text{d}^2\text{y}}{\text{dt}^2}=-\text{A}\omega^2\sin(\omega\text{t}+\phi)$

$=-\text{A}\text{y}\omega^2=(-\text{A}\omega^2)\text{y}$

$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dt}^2}\propto(-\text{y})$

Hence, it is an equation of SHM with amplitude

$\text{A}=\sqrt{\text{a}^2+\text{b}^2}$

1. The force is zero at $\text{t}=\frac{3\text{T}}{4}.$

2. The acceleration is maximum at $\text{t}=\frac{4\text{T}}{4}.$

3. The velocity is maximum at $\text{t}=\frac{\text{T}}{4}.$

Explanation:

1. At $\text{t}=\frac{3\text{T}}{4}$Particle is at it's mean position so force acting on it is zero, but it continue the motion due to inertia of mass, here a = 0, so F = 0.

2. At $\text{t}=\frac{4\text{T}}{4}=\text{T},$particle's velocity changes increasing to decreases so maximum change in velocity at T. As Acceleration$=\frac{\text{change in velocity}}{\text{Time}},$so acceleration is maximum here.

3. $\text{t}=\frac{\text{T}}{2}=\frac{2\text{T}}{4},$the particles has K.E = 0. So $\text{KE}\neq\text{PE}.$

1. $\text{x}(\text{t})=\text{B}\sin\Big(\frac{2\pi\text{t}}{30}\Big).$

Explanation:

As the particle P is executing circular motion with radius B. let particle P is at Q at instant t, foot of perpendicular on x-axis is at R vector OQ makes $<\theta$ with its zero position not p displacement for O to R.

$\text{x}=\text{OQ}\cos(90^\circ-\theta) $

$\text{x}=\text{OQ}\sin\theta=\text{B}\sin\omega\text{t}\therefore\theta=\omega\text{t}$

$\text{x}=\text{B}\sin\frac{2\pi}{\text{T}}\text{t}$

$\therefore\text{x}=\text{B}\sin\Big(\frac{2\pi}{30}\text{t}\Big)$

4. $\text{Zero}$