MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 11

A cubical block of side 'a' is floating in a fixed and closed cylindrical container of radius

$2a$ kept on the ground. Density of the block is

$\rho$ , whereas the density of liquid is

$2\rho$ . Container is made up of conducting wall so that the temperature remains constant. A piston is mounted in the cylinder which can move inside the cylinder without friction. If piston oscillates with large amplitude A.

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The cube will remain stationary
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The cube will oscillate with very small amplitude in same phase with piston
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The cube will oscillate with very small amplitude in opposite phase with piston
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The cube will oscillate with amplitude A

(b) When the block is at position

$B$ on the graph. its

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position and velocity are positive
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position is positive and velocity is negative
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position in negative and velocity is positive
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position and velocity are negative

When the block is at position

$C$ on the graph, its

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velocity is maximum and acceleration is zero
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velocity if minimum and acceleration is zero
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velocity is zero and acceleration is negative
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velocity is zero and acceleration if positive

(b) Position of the block as a function of time can now be expressed as

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$x=2\sqrt{3}\cos\left(16t+\dfrac{\pi}{6}\right)cm$
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$x=3\cos\left(16t+\dfrac{\pi}{3}\right)cm$
0%
$x=3.8\cos\left(16t+\dfrac{\pi}{6}\right)cm$
0%
$x=3.2\cos \left(16t+\dfrac{\pi}{4}\right)cm$

Explanation

$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{25600}{100}}=16\ rad/s$
Using energy conservation,

$\dfrac{1}{2}kx^{2}+\dfrac{1}{2}mv^{2}=\dfrac{1}{2}kA^{2}$

$A=\sqrt{x^{2}+\dfrac{m}{k}v^{2}}=\sqrt{9+3}=2\sqrt{3}$

$x=A\cos (\omega t+\phi)=2\sqrt{3}\cos (16t+\phi)$
From boundary conditions

$t=0, \phi$ can be obtained which is

$\pi/6$

$x=2\sqrt{3}\cos\left(16t+\dfrac{\pi}{6}\right)cm$

A particle executes SHM starting from its mean position at

$t=0$ . If its velocity is

$\sqrt 3 b\omega$ , when it is at a distance

$b$ from the mean position, when

$\omega==2\pi /T$ , the time taken by the particle to move from

$b$ to the extreme position on the same side is

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$\dfrac{5\pi}{6\omega}$
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$\dfrac{\pi}{3\omega}$
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$\dfrac{\pi}{2\omega}$
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$\dfrac{\pi}{4\omega}$

Explanation

$v^2=\omega^2 (A^2-b^2)=3\omega^2 b^2$

$\Rightarrow A^2 =4b^2$

$b=\dfrac {A}{2} =A\sin \omega t \ \Rightarrow t=\dfrac {\pi}{6\omega}$

Required time,

$t_1 =\dfrac {T}{4}-t=\dfrac {2\pi}{4\omega}-\dfrac {\pi}{6\omega}=\dfrac {\pi}{3\omega}$

A cork floating on the pond water executes a simple harmonic motion, moving up and down over a range of

$4\ cm$ . The time period of the motion is

$1\ s$ . At

$t=0$ , the cork is at its lowest position of oscillation, the position and velocity of the cork at

$t=10.5\ s$ , would be

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$2\ cm$ above the mean position, $0\ m/s$
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$2\ cm$ below the mean position, $0\ m/s$
0%
$1\ cm$ above the mean position, $2\sqrt 3 \pi \ m/s$ up
0%
$1\ cm$ below the mean position, $2\sqrt 3 \pi \ m/s$ up

Explanation

As the range of motion is

$4\ cm$ , the amplitude of motion is

$+2\ cm$ .

$10.5\ s$ is equal to

$10.5$ time period of simple harmonic motion; so we have to find the height and position of cork after one-half of time period as

$\dfrac{T}{2}=0.5\ s$ . As at

$t=10\ s$ , the particle is at its lowest position, after half a time period the cork would be at its maximum height and velocity of cork at extreme position is zero.

A simple pendulum is oscillating between extreme positions

$P$ and

$Q$ about the mean position

$O$ . Which of the the following statements are true about the motion of pendulum?

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At point $O$ , the acceleration of the bob is difference from zero.
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The acceleration of the bob is constant throughout the oscillation
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The tension in the string is constant throughout the oscillation
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The tension is maximum at $O$ and minimum at $P$ or $Q$

Explanation

Statement (a) is correct. At any position

$O$ and

$P$ or between

$O$ and

$Q$ , there are two accelerations --- a tangential acceleration

$g\sin \alpha$ and a centripetal acceleration

$v^{2}/l$ (because the pendulum moves along the arc of a circle or radius

$l$ ), where

$l$ is the length of the pendulum and

$v$ its speed at that position. When the bob is at the mean position

$O$ , the angle

$\alpha=0$ , therefore

$\sin \alpha=0$ ; hence, the tangential acceleration is zero. But at

$O$ , speed

$v$ is maximum and the centripetal acceleration

$v^{2}/l$ is directed radially towards the point of support. When the bob is at the end points

$P$ and

$Q$ the speed

$v$ is zero, hence the centripetal acceleration is zero at the end, points, but the tangential acceleration is maximum and is directed along the tangent to the curve at

$P$ and

$Q$ . The tension in the string is not constant throughout the oscillation. At any position between

$O$ and end point

$P$ or

$Q$ , the tension in the string is given by

$T=mg\cos \alpha$ .
At the end point

$P$ and

$Q$ the value of

$\alpha$ is the greatest, hence the tension is the least. At the mean position

$O$ .

$\alpha=0$ and

$\alpha=1$ which is the greatest; hence tension is greatest at the mean position.
1599131_1748643_ans_a6f1b2d12e354e3c99e62826aae19c81.png

(b) Velocity of the block as a function of time can be expressed as

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$v=-48\sin \left(16t\dfrac{\pi}{2}\right)cm/s$
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$v=-48\sin \left(16t\dfrac{\pi}{3}\right)cm/s$
0%
$v=-56\sin \left(16t\dfrac{\pi}{4}\right)cm/s$
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$v=-56\sin \left(16t\dfrac{\pi}{6}\right)cm/s$

The displacement of a particle is represented by the equation

$y = 3 \cos \left [ \dfrac{\pi }{4} - 2\omega t \right ]$ the motion of the particle is

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Simple harmonic with period $\dfrac{2\pi }{\omega }$
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Simple harmonic with period $\dfrac{\pi }{\omega }$
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Periodic but not simple harmonic.
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Non-periodic.

Explanation

As we know,

A Simple harmonic 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.

Now,

given equation is

$y=3cos[\dfrac{\pi}{4}-2\omega t]$

on comparing with

$y=A_0cos(kx-\omega ' t)$

we get,

$\omega '=2\omega$

$\dfrac{2\pi}{T}=2\omega$

$T=\dfrac{\pi}{\omega}$

Hence the motion is simple harmonic with time period

$\dfrac{\pi }{\omega }$ .
Verifies the option (b).

The displacement of a particle is represented by the equation

$y = \sin ^{3}\omega t$ The motion is

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non-periodic.
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Periodic but not simple harmonic.
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Simple harmonic with period $\dfrac{2\pi }{\omega}$
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Simple harmonic with period $\dfrac{\pi }{\omega}$

Explanation

As we know that all function of single sine and cosine function of "t" is an SHM.

But, here equation is:

$y=sin^3\omega t=\dfrac{1}{4}[3sin\omega t-sin3\omega t]$

As this motion is not represented by a single harmonic function, hence it is not SHM. As this motion involves sine and cosine functions, hence it is periodic motion.

Which of the following statements are true for a stationary wave?

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Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.
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All the particles cross their mean position at the same time.
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All the particles are oscillating with same amplitude.
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There is no net transfer of energy across any plane.
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There are some particles which are always at rest.

Explanation

Standing waves have certain properties:

(i). Every particle has a fixed amplitude. It is different for different particles.

(ii) The particle cross their mean position in same time

(iii)The energy becomes alternately whole potential and kinetic twice in each cycle. There is no net transfer of energy.

(iv) Some particles are in rest always. They are called

$nodes$

The correct answers are

$A,B,D,E$

Which of the following statements is /are true for a simple harmonic oscillator?

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Force acting is directly proportional to displacement from the mean position and opposite to it.
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Motion is periodic.
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Acceleration of the oscillator is constant.
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The velocity is periodic.

Explanation

Comsider a SHM

$x = a \sin \omega t. ....(i)$

$v = \dfrac{dx}{dt} = a \omega \cos \omega t(ii)$

Acceleration

$A = \dfrac{dv}{dt} = d\dfrac{d}{dt}(a \omega \cos \omega t) = a \omega(\omega \sin \omega t)$

A ball thrown by a boy from a roof-top has oscillatory motion.

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True
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False

Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is

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Simple harmonic motion.
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Non-periodic motion.
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Periodic motion.
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Periodic but not SHM

The displacement time graph of a particle executing S.H.M. is shown in figure Which of the following statement is/are true?

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The force is zero at $t = \dfrac{3 T}{4}$
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The acceleration is maximum at $t = \dfrac{4 T}{4}$
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The velocity is maximum at $t = \dfrac{T}{4}$
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The P.E. is equal to K.E. of oscillation at $t = \dfrac{T}{2}$

The relation acceleration and displacement of four particles are given below.
Which one particle is executing simple harmonic motion?

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$a_{x} = + 2x$
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$a_{x} = 2x^{2}$
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$a_{x} = - 2x^{2}$
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$a_{x} = - 2x$

Explanation

As we know,

In simple harmonic motion, force is proportional and opposite to displacement.

$F=-kx$

Applying the equation of motion

$F = Ma,$

$a=-kx$
Verifies the option (d).

Figure shows the circular motion of a particle. The radius of the circle, The period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is

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$x(t) = B \sin \left ( \dfrac{2 \pi t}{30} \right )$
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$x(t) = B \cos \left ( \dfrac{\pi t}{15} \right )$
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$x(t) = B \sin \left ( \dfrac{\pi t}{15} + \dfrac{\pi }{2} \right )$
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$x(t) = B \cos \left ( \dfrac{\pi t}{15} + \dfrac{\pi }{2} \right )$

Explanation

$x(t)= A sin \dfrac {2\pi t}{T}$ or

$A cos \dfrac {2 \pi t}{T}$

At t=0

$x(t)=0 or A$

So, the first option satisfies as the Simple harmonic equation.

Two masses

$m_1$ and

$m_2$ are suspended together by a massless springs of constant

$k$ . When the masses are in equilibrium,

$m_1$ is removed without disturbing the system. Then the angular frequency of oscillation of

$m_2$ is

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$\sqrt{\dfrac{k}{m_1}}$
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$\sqrt{\dfrac{k}{m_2}}$
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$\sqrt{\dfrac{k}{m_1+m_2}}$
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$\sqrt{\dfrac{k}{m_1m_2}}$

Explanation

When only the mass

$m_2$ is suspended let the elongation of the spring be

$x_1$ . When both the masses

$(m_2 + m_1)$ together are suspended, the elongation of the spring is

$(x_1 + x_2)$ . Thus, we have

$m_2g=kx_1$

and,

$(m_1+m_2)g=k(x_1+x_2)$ , where k is the spring constant.

$\Rightarrow m_1g+m_2g =kx_1 + kx_2$

$\Rightarrow m_1=kx_2$ , since

$m_2g=kx_1$ .

Thus,

$x_2$ is the elongation of the spring due to the mass

$m_1$ only. When the mass

$m_1$ is removed the mass

$m_2$ executes SHM with the amplitude

$x_2$ .

Amplitude of vibration,

$x_2 = m_1g$

Hence, angular frequency,

$\omega = \sqrt{\frac{k}{m_2}}$ is our required answer.

What is constant in S.H.M.

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Restoring force
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Kinetic energy
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Potential energy
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Periodic time

A system exhibiting

$S.H.M$ must possess

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Inertia only
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Elasticity as well as inertia
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Elasticity inertia and an external force
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Elasticity only

Which of the following is not true? In the case of a simple pendulum for small amplitudes the periodic of oscillation is

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Directly proportional to square root of the length of the pendulum
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Inversely proportional to square root of the acceleration due to gravity
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Dependent on the mass, size and material of the bob
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Independent of the amplitude

Explanation

Time period is given by,

$T=2\sqrt{\frac{l}{g}}$

Clearly,

$T\propto \sqrt{l}$ . So, option A is correct.

Also,

$T\propto\frac{1}{\sqrt{g}}$ So, option B is also correct.

We don't see any amplitude term in the time period equation hence, its independent of amplitude. So, option D is also correct.

We don't see any mass or size or material dependency in the time period equation. So, its not dependent upon mass or size or material of the bob. Hence, option C is the wrong statement here.

A mass

$m$ is suspedend from the two coupled springs connected in series. The force constant for springs are

$K_1$ and

$K_2$ . The time period of the suspended mass will be

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$T=2\pi \sqrt{\left(\dfrac{m}{K_1+K_2}\right)}$
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$T=2\pi \sqrt{\left(\dfrac{m}{K_1+K_2}\right)}$
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$T=2\pi \sqrt{\left(\dfrac{m(K_1+K_2)}{K_1K_2}\right)}$
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$T=2\pi \sqrt{\left(\dfrac{mK_1K_2}{K_1+K_2}\right)}$

Explanation

In Series

$K_{eq}=\dfrac{K_1K_2}{K_1+K_2}$

so Time period

$T=\sqrt{\dfrac{m}{K_{eq}}}=2\pi \sqrt{\dfrac{m(K_1+K_2)}{K_1K_2}}$

A simple pendulum is sec up in a trolley which moves to the right with an acceleration

$a$ on a horizontal plane. Then the thread of the pendulum in the mean position makes an angle

$\theta$ with the vertical

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$\tan^{-1} \dfrac ag$ in the forward direction
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$\tan^{-1} \dfrac ag$ in the backward direction
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$\tan^{-1} \dfrac ga$ in the backrward direction
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$\tan^{-1} \dfrac ga$ in the forward direction

Explanation

In acceleration frame of reference a frictions force (pseudo force)

$ma$ acts on the bob of pendulum as shown in figure
Hence,

$\tan \theta =\dfrac {ma}{mg}=\dfrac ag$

$\Rightarrow$

$\theta =\tan^{-1}\left(\dfrac ag \right)$ in the backward direction.
1727947_1813625_ans_ef88bca046fe4e04a9d39587289fe785.png

The period of a simple pendulum is doubled, when

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Its length is doubled
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The mass of the bob is doubled
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Its length is made four times
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The mass of the bob and the length of the pendulum are doubled

A particle moving along the

$x-$ axis execute simple harmonic motion, then the force acting on it is given by

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$-A\ Kx$
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$A\ \cos \ (Kx)$
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$A\ exp \ (-Kx)$
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$A\ Kx$

Explanation

For SHM, we need

$F\propto -x$ , which means that force will act in the opposite direction to force.

Looking at the given options we see that only

$F=-AKx$ will be satisfying the condition of SHM if we assume

$AK$ to be the constant of proportionality.

The period of simple pendulum is measured as

$T$ in a stationary lift. If the lift moves upward with an acceleration of

$5\ g$ , the period will be

0%
The same
0%
Increased by $3/5$
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Decreased by $2/3$ times
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None of the above

Explanation

$T=2\pi\sqrt{\dfrac{l}{g}}\\T\propto \sqrt{\dfrac{1}{g}}$

$\dfrac {T'}{T}=\sqrt {\dfrac {g}{g+a}}=\sqrt {\dfrac {g}{g+5g}}=\sqrt {\dfrac {1}{6}}$

$\Rightarrow T' =\dfrac {T}{\sqrt 6}$

Mark the wrong statement

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All S.H.M.s have fixed time period
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All motion having same time period are S.H.M.
0%
In S.H.M. total energy is proportional to square of amplitude
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Phase constant of S.H.M. depends upon initial conditions

Explanation

For SHM, the force of the particle is given by,

$F=-kx$

$\Rightarrow ma=-kx$

$\Rightarrow a=\frac{-k}{m} x$

We also have,

$a=-\omega^2 x$
From this, we can say

$\omega^2=\dfrac{k}{m}$

or,

$\omega^2=\sqrt{\dfrac{k}{m}}$ , which is a constant.

Now, time period is given by,

$T=\dfrac{2\pi}{\omega}$ which will also be a constant. Hence, option A is right.

All motion which have same period may not be SHM. A good example is that of when a particle is moving in a circle. This case has same time period but is not an example of SHM. Hence, the wrong answer is B.

The total energy of SHM is given by,

$KE=\dfrac{1}{2}kA^2$ . So, we can see that the total energy is directly proportional to the square of amplitude. Hence, option C is also correct.

The phase constant is given by

$y=Asin\omega t + \phi$ where

$\phi$ is the phase constant. Clearly we can see that the phase constant depends directly upon the initial condition of the particle. Hence, option D is also correct.

The time period of a simple pendulum is

$2$ sec. If its length is increased

$4$ times, then its period becomes

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$16$ sec
0%
$12$ sec
0%
$8$ sec
0%
$4$ sec

A block is placed on a frictionless horizontal table. The mass of the block is

$m$ and springs are attached on either side with force constants

$K_1$ and

$K_2$ . If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be

0%
$\left( \dfrac{K_1+K_2}{m}\right)^{1/2}$
0%
$\left[ \dfrac{K_1K_2}{m(K_1+K_2)}\right]^{1/2}$
0%
$\left[ \dfrac{K_1K_2}{(K_1-K_2)m}\right]^{1/2}$
0%
$\left[ \dfrac{K_1^2+K_2^2}{(K_1+K_2)m}\right]^{1/2}$

Explanation

In this case springs are parallel, so

$k_{eq}=k_1+k_2$

and

$\omega =\sqrt{\dfrac{k_{eq}}{m}}=\sqrt{\dfrac{k_1+k_2}{m}}$

In a simple pendulum, the period of oscillation

$T$ is released to length of the pendulum

$l$ as

0%
$\dfrac {l}{T}=$ constant
0%
$\dfrac {l^2}{T}=$ constant
0%
$\dfrac {l}{T^2}=$ constant
0%
$\dfrac {l^2}{T^2}=$ constant

Graph between velocity and displacement of a particle, executing S.H.M. is

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A straight line
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A parabola
0%
A hyperbola
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An ellipse

Explanation

We know that the displacement function of a particle performing SHM is given by,

$y=Asin(\omega t+\phi)$

or,

$sin(\omega t+\phi)=\frac{y}{A}$ (1)

The velocity can be found by differentiating the displacement function with respect to time,

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

or,

$cos(\omega t+\phi)= \frac{v}{A\omega}$ (2)
Squaring and adding equations (1) and (2) we get,

$\frac{y^2}{A^2}+\frac{v^2}{(A\omega)^2}=1$ (

$\because sin^2\theta +cos^2\theta =1$ )

This equation resembles the equation of a ellipse.

(equation of ellipse is

$\frac{x^2}{a^2} + \frac{y^2}{b^2} =1$ )

If the length of simple pendulum is increased by

$300\%$ ,then the time period will be increased by

0%
$100\%$
0%
$200\%$
0%
$300\%$
0%
$400\%$

A simple pendulum is executing simple harmonic motion with a time period

$T$ . If the length of the pendulum is increased by

$21\%$ , the percentage increase in the time period of the pendulum of increased length is

0%
$10\%$
0%
$21\%$
0%
$30\%$
0%
$50\%$

Explanation

If initial length

$l_1=100$ then

$l_2=121$
By using

$T=2\pi \sqrt {\dfrac {l}{g}}\Rightarrow \dfrac {T_1}{T_2}=\sqrt {\dfrac {l_1}{l_2}}$

Hence,

$\dfrac {T_1}{T_2}=\sqrt {\dfrac {100}{121}}\Rightarrow T_2 =1.1 T_1$

$\%$ increase

$=\dfrac {T_2 -T_1}{T_1}\times 100=10\%$

The force constants of two springs are

$K_1$ and

$K_2$ . Both are stretched till their elastic energies are equal. If the stretching forces are

$F_1$ and

$F_2$ , then

$F_1 : F_2$ is

0%
$K_1: K_2$
0%
$K_2: K_1$
0%
$\sqrt{K_1}: \sqrt{K_2}$
0%
$K_1^2: K_2^2$

Explanation

Given elastic energies are equal

i.e.

$\dfrac 12 k_1x_1^2=\dfrac 12 k_2x_2^2$

$\Rightarrow \dfrac{k_1}{k_2}=\left( \dfrac{x_2}{x_1}\right)^2$ and using

$F=kx$

$\Rightarrow \dfrac{F_1}{F_2}=\dfrac{k_1x_1}{k_2x_2}=\dfrac{k_1}{k_2}\times \sqrt{\dfrac{k_2}{k_2}}=\sqrt{\dfrac{k_1}{k_2}}$

If a simple harmonic oscillator has got a displacement of

$0.02\ m$ and acceleration equal to

$2.0ms^{-2}$ at any time, the angular frequency of the oscillator is equal to

0%
$10\ rad\ s^{-1}$
0%
$0.1\ rad\ s^{-1}$
0%
$100\ rad\ s^{-1}$
0%
$1\ rad\ s^{-1}$

Explanation

When a particle undergoes SHM, its acceleration is given by,

$\alpha = \omega^2 x$

Given,

$\alpha = 2, x=0.02$ . Using these values

$\omega =\sqrt{\frac{\alpha}{x}}$

or,

$\omega=\sqrt{\frac{2}{0.02}}$

or,

$\omega=10\space rad/s$ is our required answer.

What will be the force constant of the spring system shown in the figure

0%
$\dfrac{K_1}{2}+K_2$
0%
$\left[ \dfrac{1}{2K_1}+\dfrac{1}{K_2}\right]^{-1}$
0%
$\dfrac{1}{2K_1}+\dfrac{1}{K_2}$
0%
$\left[ \dfrac{2}{K_1}+\dfrac{1}{K_1}\right]^{-1}$

Explanation

In series combination

$\dfrac{1}{k_s}=\dfrac{1}{2k_1}+\dfrac{1}{k_2}$

$\Rightarrow$

$k_S=\left[ \dfrac {1}{2k_1}+\dfrac{1}{k_2}\right]^{-1}$
1729706_1813698_ans_3079b0c2d5964629960e6a3fbd9451d7.png

A simple pendulum oscillates in air with time period

$T$ and amplitude

$A$ . As the time passes

0%
$T$ and $A$ both decrease
0%
$T$ increases and $A$ is constant
0%
$T$ increases and $A$ decreases
0%
$T$ decreases and $A$ is constant

Two simple pendulum of length

$1.44\ m$ and

$1\ m$ start swinging together. After how many vibrates will they again start swinging together

0%
$5$ oscillations of smaller pendulum
0%
$6$ oscillations of smaller pendulum
0%
$4$ oscillation of bigger pendulum
0%
$6$ oscillation of bigger pendulum

Explanation

$n\propto \dfrac{1}{\sqrt{l}}$

$\Rightarrow \dfrac{n_{2}}{n_{1}}=\sqrt{\dfrac{1.44}{1}}=\dfrac{1.2}{1}$

$\Rightarrow n_{2}=1.2\ n_{1}$

For

$n_2$ be integer minimum value of

$n_1$ should be

$5$ and then

$n_2=6$ i.e., after

$6$ oscillations of smaller pendulum both will be in phase.

A simple pendulum is set into vibrations. The bob of the pendulum comes to rest after some time due to

0%
Air friction
0%
Moment of inertia
0%
Weight of the bob
0%
Combination of all the above

The velocity of simple pendulum is maximum at

0%
Extremes
0%
Half displacement
0%
Mean position
0%
Every where

The time period of simple pendulum when it is made to oscilate on the surface of moon

0%
Increase
0%
Decrease
0%
Remains changed
0%
Becomes infinite

Explanation

$T=2\pi\sqrt{\dfrac{l}{g}}$

$\left(as\ T \propto \dfrac {1}{\sqrt g}\right)$

At the surface of moon,

$g$ decrease hence time period increase

On a planet a freely falling body takes

$2\ sec$ when it is dropped from a height of

$8\ m$ , the time period of simple pendulum of length

$1\ m$ on that planet is

0%
$3.14\ sec$
0%
$6.28\ sec$
0%
$1.57\ sec$
0%
None of these

Explanation

On a planet, if a body dropped initial velocity

$(u = 0)$ from a height

$h$ and takes time

$t$ to reach the ground then

$h=\dfrac{1}{2}g_{P}t^{2}\Rightarrow g_{P}=\dfrac{2\ h}{t^{2}}=\dfrac{2\times 8}{4}=4\ m/s^{2}$

Using

$T=2\pi \sqrt{\dfrac{l}{g}}\Rightarrow T=2\pi \sqrt{\dfrac{1}{4}}=\pi =3.14\ sec$

A particle on the trough of a wave at any instant will come to the mean position after a time (t = time period) [KCET 2005]

0%
T/2
0%
T/4
0%
T
0%
2T

Explanation

The particle will come after a time

$\frac{T} {4}$ to its mean position.

The graph shows variation of displacement of a particle performing S.H.M. with time t. which of the following statements is correct from the graph ?

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The acceleration is maximum at time T .
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The force is maximum at time $\frac { 3T}{4}$
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The velocity is zero at time $\frac { T}{2 }$
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The kinetic energy is equal to total energy at a time $\frac {T}{4}$

A set of keys on the end of a string is swung steadily in a horizontal circle. In one trial, it moves at speed

$v$ in a circle of radius

$r$ . In a second trial, it moves at a higher speed

$4v$ in a circle of radius

$4r$ . In the second trial, how does the period of its motion compare with its period in the first trial?

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It is the same as in the first trial.
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It is $4$ times larger.
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It is one-fourth as large.
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It is $16$ times larger.
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It is one-sixteenth as large.

In the above question, the velocity of the rear 2 kg block after it separates from the spring will be :

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0 m/s
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5 m/s
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10 m/s
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7.5 m/s

A mass

$m$ is undergoing SHM in the vertical direction about the mean position

$y_{0}$ with amplitude A and angular frequency

$\omega$ . At a distance

$y$ from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of

$m$ . Find the distance

$y^{*}$ (measured from the mean position) such that the height

$h$ attained by the block is maximum.

$(\mathrm{A}\mathrm{\omega})^{2}>g$

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$\dfrac{g}{\omega^2}$
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$\dfrac{2g}{\omega^2}$
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$\dfrac{g}{2\omega^2}$
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None of these

Explanation

Let

$v$ be the upward velocity of the block when it detaches from the spring at a distance

$y$ above the mean position.

$v = w \sqrt{A^2 - y^2} \implies v^2 = w^2 (A^2 - y^2)$ ..........(1)

At the detaching point, the block will continue to move upwards due to inertia and reach the height

$h$ above the mean position. At that height the velocity of the block will be zero.

Thus

$0 - v^2 = 2(-g)s \implies s = \dfrac{v^2}{2g}$

Now the total height

$h =s + y = \dfrac{v^2}{2g} + y$

$\therefore h = \dfrac{w^2 (A^2 - y^2)}{2g} + y$

For

$h$ to be maximum,

$\dfrac{dh}{dy} = 0$

Thus

$\dfrac{w^2}{2g} (-2 y) + 1 = 0$

$\implies y = \dfrac{g}{w^2}$

A particle of mass m moves due to a conservative force with potential energy V(x) =

$\frac{Cx}{x^{2}+a^{2}}$ , where C and a are positive constants. The position(s) of stable equilibrium is/are given as

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$x = + a$ only
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$x = - a$ only
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$x = - \dfrac{a}{2}$ and $+ \dfrac{a}{2}$
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$x = - a$ and $+ a$

Explanation

At equilibrium position

$\displaystyle \mathrm{F}=-\dfrac{dV}{dx}=0$

$\dfrac{dV}{dx }=\dfrac{C(a^{2}-x ^{2})}{(x ^{2}+a^{2})^{2}}=0$

$\therefore$ There are two equilibrium positions

$x_1 = a$

$x_2 = -a$

Consider $\dfrac{d^{2}V}{dx ^{2}}=\dfrac{2Cx (x ^{2}-3a^{2})}{(x^{2}+a^{2})^{3}}$

$\dfrac{d^{2}V}{dx ^{2}}|_{x_1}< 0$

$\dfrac{d^{2}V}{dx ^{2}}|_{x_2}> 0$

$\because$ There is a maxima at $x=a$ and minima at $x=-a$
$\Rightarrow x _{1}$ is a position of unstable equilibrium and $x _{2}$ is a position of stable equilibrium.

A particle of mass

$m$ is bound by a linear potential

$U = Kr$ . It will have stationery circular motion with angular frequency

$\omega _{o}$ with radius

$r$ about the origin. If the particle is slightly disturbed from this circular motion, it will have small oscillations. If the angular frequency

$\omega$ of the oscillations is

$\omega=\sqrt{n}\omega _{o}$ The value of

$n$ is :

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$2$
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$3$
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$5$
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$6$

Explanation

$E_{total}$ = U + (kinetic energy)

$=Kr+\dfrac{mv^{2}}{2}=Kr+\dfrac{m\omega ^{2}r^{2}}{2}$

$=\dfrac{3Kr}{2}$

(

$\because$ for circular motion F = m

$\omega ^{2}r$

$=\left | \dfrac{dU}{dr} \right |=K$ )

Angular momentum about the origin:

L =

$m\omega r^{2}$

$=mr^{2}\sqrt{\dfrac{K}{mr}}$

$=\sqrt{mKr^{3}}$

Angular frequency of circular motion is

$\omega =\sqrt{\dfrac{K}{mr}}$

Effective potential

$U_{eff}=Kr+\dfrac{L^{2}}{2mr^{2}}$

Radius

$r_{o}$ of the stationery circular motion is

$\left ( \dfrac{dU_{eff}}{dr} \right )_{r=r_{o}}=K-\dfrac{L^{2}}{mr_{0}^{3}}=0$

$\Rightarrow r_{o}=\left ( \dfrac{L^{2}}{mK} \right )^{\frac{1}{3}}$

$\left ( \dfrac{d^{2}U_{eff}}{dr^{2}} \right )_{r=r_{o}}=\dfrac{3L^{2}}{mr^{4}}|_{r=r_{o}}$

$=\dfrac{3L^{2}}{m}\left ( \dfrac{mK}{L^{2}} \right )^{\frac{4}{3}}$

The angular frequency of small radial oscillations about

$r_{o},$ if it is slightly disturbed from the stationary circular motion:

$\omega _{r}=\sqrt{\dfrac{1}{m}\left ( \dfrac{d^{2}U_{eff}}{dr^{2}} \right )}_{r=r_{o}}$ =

$\sqrt{\dfrac{3K}{m}\left ( \dfrac{mK}{L^{2}} \right )}^{\frac{1}{3}}$

$=\sqrt{\dfrac{3K}{mr_{o}}}=\sqrt{3}\omega _{o}$ , where

$\omega _{o}$ is angular frequency of stationary circular motion.

So, according to given condition,

$n=3$ .

A particle of mass m moves under a conservative force with potential energy.

$V(x)=\dfrac{ Cx}{\sqrt{a^2+x^2}, }$ where

$C$ and

$a$ are positive constants.

If the practicle starts from a point with velocity

$v$ , the range of values of

$v$ for which it escapes to -

$\infty$ are given by

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v $< \sqrt{\dfrac{C}{ma}}$
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v $> \sqrt{\dfrac{C}{ma}}$
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v $> \sqrt{\dfrac{2C}{ma}}$
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v $< \sqrt{\dfrac{2C}{ma}}$

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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