Physics M.C.Q (1 Marks)

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. 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.

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.

3. Simple harmonic and time period is independent of the density of the liquid.

Explanation:

Restoring Force F = Weight of liquid column of height 2y

$\Rightarrow\text{FF}=-(\text{A}\times2\text{y}\times\rho)\times\text{g}=-2\text{A}\rho\text{gy}$

$\Rightarrow\text{F}\propto-\text{y}$

Motion is SHM with force constant

$\text{K}=2\text{A}\rho\text{g}$

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

$=2\pi\sqrt{\frac{\text{A}\times2\text{h}\times\rho}{2\text{A}\rho\text{g}}}=2\pi\sqrt{\frac{\text{h}}{\text{g}}}$

Which is independent of the density of the liquid.

2. Phase of the oscillator is same at t = 2 s and t = 6 s.

4. Phase of the oscillator is same at t = 1 s and t = 5 s.

Explanation:

Two particles are said to be in same phases if the mode of vibration is same i.e, their distance will be $\text{n}\lambda(/\text{n}=1,2,3...)$

Distance between particles at t = 0 and t = 2 is $\frac{\lambda}{2}.$

So, articles are not in same phase.

As from figure the particles at t = 2 sec are at distance $\lambda$, so are in same phase.

Particles at t = 1, t = 7 are the distance $\lambda+\frac{\lambda}{2}=\frac{3\lambda}{2}$ so are not in phase.

Particles at t = 1 and 5sec are at distance = $\lambda$ so are in same phase.

1. Force acting is directly proportional to displacement from the mean position and opposite to it.

2. Motion is periodic.

4. The velocity is periodic.

Explanation:

Let us write the equation for the SHM $\text{x}=\text{a}\sin(\omega\text{t}+\phi)$

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

Let us find velocity of the particle, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin(\omega\text{t}+6\phi))$

Velocity is also periodic because it is a cosine function.

Now let us find acceleration, $\text{A}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{a}\omega^2\sin(\omega\text{t}+\phi)$

The acceleration is a sine function, hence cannot be constant.

$\Rightarrow\text{A}=-(\omega^2\text{a})\sin(\omega\text{t}+\phi)=-\omega^2\text{x}$

Force, F = mass × acceleration.

$=\text{mA}=-\text{m}\omega^2\text{x}$

Hence, force acting is directly proportional to displacement from the mean position and opposite to it.

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

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) $

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

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

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

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 $\frac{\pi}{\omega}.$

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. $\sqrt{\text{v}_1^2+\text{v}_2^2}.$

Explanation:

When the mass is connected to the two springs individually.

$\text{v}_1=\frac{1}{2\pi}\sqrt{\frac{\text{k}_1}{\text{m}}}\ ...(1)$

$\text{v}_2=\frac{1}{2\pi}\sqrt{\frac{\text{k}_2}{\text{m}}}\ ...(2)$

Now, the block is connected with two springs considered as parallel.

Here equivalent spring constant $\text{K}_\text{eq}=\text{K}_1+\text{K}_2$

Time period of oscillation of the spring block-system is

$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}_\text{eq}}}=2\pi\sqrt{\frac{\text{m}}{\text{k}_1+\text{k}_2}}$

Hence frequency,

$\text{v}=\frac{1}{\text{T}}=\frac{1}{2\pi}\times\sqrt{\frac{\text{k}_1+\text{k}_2}{\text{m}}}\ ...(3)$

$\text{v}=\frac{1}{2\pi}\Big[\frac{\text{k}_1}{\text{m}}+\frac{\text{k}_2}{\text{m}}\Big]^{\frac{1}{2}}$

From Eq. (i) $\frac{\text{k}_1}{\text{m}}=4\pi^2\text{v}_1^2$ and from Eq.(ii) $\frac{\text{k}_2}{\text{m}}=4\pi^2\text{v}_2^2$

$\Rightarrow\text{v}=\frac{1}{2\pi}\Big[\frac{4\pi^2\text{v}_1^2}{1}+\frac{4\pi^2\text{v}_2^2}{1}\Big]^\frac{1}{2}=\frac{2\pi}{2\pi}[\text{v}_1^2+\text{v}_2^2]^\frac{1}{2}$

$\Rightarrow\text{v}=\sqrt{\text{v}_1^2+\text{v}_2^2}$

1. Oscillatory but not periodic.

Explanation:

$\text{x}=\text{a}\cos(\propto\text{t})^2$ is a cosine function and x varies between -a and +a, the motion is oscillatory. Now checking for periodic motion, putting t + T in place of t. T is supposed as period of the function ω(t).

$\text{x}(\text{t}+\text{T})=\text{a}\cos[\alpha(\text{t}+\text{T})]^2$

$=\text{a}\cos[\alpha^2\text{t}^2+\text{a}^2\text{T}^2+2\alpha^2\text{tT}]\neq\text{x}(\text{t})$

Hence, it is not periodic.

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

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

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

Explanation:

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.

$\text{Time period}=\text{T}'=\frac{2\pi}{\omega'}$

hence the motion is simple harmonic with time period $\frac{2\pi}{\omega}.$

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,

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

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

3. A circle.

Explanation:

Resultant displacement is x + y

$\text{x}=\text{a}\cos\omega\text{t}\ ...(1)$

$\text{y}=\text{A}\sin\omega\text{t}\ ...(2)$

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

$\text{y}'=\text{a}\sqrt{2}\Big[\frac{\cos\omega\text{t}}{\sqrt{2}}+\frac{\sin\omega\text{t}}{\sqrt{2}}\Big]$

$\text{y}'\text{a}\sqrt{2}\ [\cos\omega\text{t}\cos45^\circ+\sin\omega\text{t}\sin45^\circ]$as the particle is acted simultaneously by Mutually perpendicular direction.

$\text{y}'=\text{a}\sqrt{2}\cos\text{s}(\omega\text{t}-45^\circ)$

Hence the displacement is neither a straight line nor a parabola.

Now, squaring and adding (i), (ii)

$\text{x}^2+\text{y}^2=\text{a}^2\cos^2\omega\text{t}+\text{a}^2\sin^2\omega\text{t}=\text{a}^2[\cos^2\omega\text{t}+\sin^2\omega\text{t}]$

$\text{x}^2+\text{y}^2=\text{a}^2$

This shows the equation of circle, hence the motion is circular motion.