MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 3

A particle is executing simple harmonic motion of amplitude $A$ . At a distance $x$ from the centre, particle moving towards the extreme position received a blow in the direction of motion which instantaneously doubles the velocity. Its new amplitude will be

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$A$
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$\sqrt{A^{2}-x^{2}}$
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$\sqrt{2A^{2}-3x^{2}}$
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$\sqrt{4A^{2}-3x^{2}}$

Explanation

Velocity of particle in SHM is given as

$v=\omega\sqrt{A^2-x^2}\ -(1)$

Let new amplitude be

$A_1$ ,

as the velocity is doubled at distance

$x$ from the center, with angular frequency

$\omega$ remaining constant.

So,

$2v=\omega\sqrt{A_1^2-x^2}\ -(2)$

From

$(1)\ and\ (2)$ ,

$A_1=\sqrt{4A^2-3x^2}$

The period of small oscillations is

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$\pi a\sqrt{\frac{ma}{C}}$
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$\pi a\sqrt{\frac{2ma}{C}}$
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$2\pi \sqrt{\frac{2ma^{3}}{C}}$
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$2\pi a\sqrt{\frac{ma}{C}}$

Explanation

For small oscillations about the positions of stable equilibrium, let

$\mathrm{x}=-\mathrm{a}+\mathrm{x}'$ where

${\times }'< < a.$ Then the equation of motion becomes

$m\frac{d^{2}{\times }'}{dt^{2}}=\frac{-C{\times }'(2a-{\times }')}{\left [ ({\times }'-a)^{2}+a^{2} \right ]^{2}}$

$\simeq -\frac{C{\times }'}{2a^{3}}$

About x = - a : period T =

$\frac{2\pi }{\omega }=2\pi \sqrt{\frac{2ma^{3}}{C}}$

A particle of mass m moves with the potential energy U shown above.The period of the motion when the particle has total energy E is

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${2\pi\sqrt{m/k}+4\sqrt{2E/mg^{2}}}$
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${2\pi\sqrt{m/k}}$
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${\pi\sqrt{m/k}+2\sqrt{2E/mg^{2}}}$
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${2\sqrt{2E/mg^{2}}}$

Explanation

As seen from graph, we can infer that for $x<0$ particle is in SHM as in spring. and for $x>0$ particle is in influence of gravity i.e. thrown upwards with some initial velocity.
So we can divide the time period in two parts. first $T_1$ and second $T_2$
$T_1$ is half of the time period of a full SHM i.e. $T_1=\pi \sqrt{\dfrac{m}{k}}$
$E=\dfrac{1}{2}m{v_{max}}^2 \\ \Rightarrow v_{max}=\sqrt{\dfrac{2E}{m}}$
and $T_2$ is the time during which the particle remains in air when it is thrown upwards with velocity $v_{max}=\sqrt{\dfrac{2E}{m}}$
$0=\sqrt{\dfrac{2E}{m}} t - \dfrac{1}{2}gt^2$ Since $s=ut - \dfrac{1}{2}gt^2$
$\Rightarrow T_2={2\sqrt{2E/mg^2}}$
So total time time period $T=\pi \sqrt{\dfrac{m}{k}}+{2\sqrt{\dfrac{2E}{mg^2}}}$

The oscillators that can be described in terms of sine or cosine functions are called

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simple harmonic
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natural
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sympathetic
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free

Explanation

In simple harmonic motion, displacement of a particle is directly proportional to restoring force.

$F \propto -x \Rightarrow F=-kx \Rightarrow m\dfrac{d^2x}{dt^2}=-kx$ .
The solution of the above equation can be given by

$x=c_1\cos(\omega t)+c_2\sin(\omega t) = A \sin (\omega t -\phi)$ , where

$A=\sqrt{c_1^2+c_2^2}$ ,

$\omega=\sqrt{\dfrac{k}{m}}$ and

$\phi =\tan^{-1}(\dfrac{c_2}{c_1})$

$c_1$ and

$c_2$ are constants and values can be determined from the initial boundary conditions.
So simple harmonic motions can be described as sin or cosine functions.

Ans:

$A$

The time taken by a particle performing

$SHM$ to pass from point

$A$ to

$B$ where its velocities are same is

$2$ seconds.After another

$2$ seconds it returns to

$B$ . The time period of oscillation is (in seconds)

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$2\ s$
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$8\ s$
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$6\ s$
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$4\ s$

Explanation

the Velocity

$v$ in SHM is given by

$v=\omega \sqrt { { A }^{ 2 }-{ x }^{ 2 } }$ . if at two points the velocity is equal that means, they are at equal distance from mean position. if one is at

$x$ then another is at

$-x$ . So if it takes 2 seconds to reach from

$A$ to

$B$ , it will take

$1s$ for the particle to reach from

$O$ to

$B$ . i.e.

$t_{OB}=1s$
Now it is given that it takes

$2s$ to reach back at

$B$ .

$t_{BXB}=2s$
And it will take one more second to reach back at mean position.

$t_{BO}=1s$
So it will take

$t_{OB}+t_{BXB}+t_{BO}=4s$ to complete the half cycle. so it will

$8s$ to complete the full cycle.

A force

$F=-10x+2$ acts on a particle of mass

$0.1$ kg, where '

$k$ ' is in m and

$F$ in newton.If it is released from rest at

$x=-2$ m,find:(a) amplitude; (b) time period; (c) equation of motion.

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(a) ${\displaystyle\frac{11}{5}m}$ (b) ${\displaystyle\frac{\pi}{5}sec}$ (c) $x=0.2-\frac{11}{5}cos\omega{t}$
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(a) ${\displaystyle\frac{11}{3}m}$ (b) ${\displaystyle\frac{\pi}{5}sec}$ (c) $x=0.2-\frac{11}{5}cos\omega{t}$
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(a) ${\displaystyle\frac{11}{5}m}$ (b) ${\displaystyle\frac{\pi}{3}sec}$ (c) $x=0.2-\frac{11}{5}cos\omega{t}$
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(a) ${\displaystyle\frac{11}{5}m}$ (b) ${\displaystyle\frac{\pi}{5}sec}$ (c) $x=0.2-\frac{11}{3}cos\omega{t}$

Explanation

Force,

$F=-10x+2$

$F=-10(x-\dfrac{1}{5})$

Let

$y=x-\dfrac{1}{5}$ ,

$F=-10y$

Now

$F=-kx$

So,

$k=10N/m$

a) Amplitude = Maximum displacement

$=|y| =|x-\dfrac{1}{5}| = |(-2)-\dfrac{1}{5}| = \dfrac{11}{5}$

b) Now,

$\omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{10}{0.1}}=10 rad/s$

Time period,

$T=\dfrac{2\pi}{\omega} = \dfrac{2\pi}{10} =\dfrac{\pi}{5} sec$

c) Finally, equation of motion will be given by:

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

Now at

$t=0$ when displacement is maximum.

$-\dfrac{11}{5} = \dfrac{11}{5}cos\phi$

So,

$\phi = cos^{-1}(-1) = \pi$

Therefore,

$y=Acos(\omega t +\pi) = -Acos(\omega t)$

Substitute for y in terms of x

$x-\dfrac{1}{5}=-\dfrac{11}{5}cos(\omega t)$

$x=\dfrac{1}{5} - \dfrac{11}{5}cos(\omega t)$

The acceleration-displacement (a-x) graph of a particle executing simple harmonic motion is shown in the figure. Find the frequency of its oscillation.

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${\displaystyle\frac{1}{2\pi}\sqrt\frac{2\beta}{ \alpha}}$
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${\displaystyle\frac{1}{2\pi}\sqrt\frac{\beta}{2\alpha}}$
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${\displaystyle\frac{1}{2\pi}\sqrt\frac\beta \alpha}$
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${\displaystyle\frac{1}{2\pi}\sqrt\frac{\beta} {\alpha - \beta}}$

Explanation

Answer: 1
For Simple Harmonic Motion,

$a(t)=-\omega^2 x(t)$
Differentiate with respect to x.

$\frac{da}{dx}=-\omega^2\frac{dx}{dx}=-\omega^2$
From graph given the slope is

$-\frac{\beta}{\alpha}$
Therefore,

$\frac{da}{dx}=-\frac{\beta}{\alpha}$

$\omega=\sqrt{\frac{\beta}{\alpha}}$
Finally, frequency of oscillation,

$f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{\beta}{\alpha}}$

Speed v of a particle moving along a straight line,when it is at a distance x from a fixed point on the line is given by

${v^{2}=108 -9x^{2}}$ (all quantities in S.I unit).Then

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The motion is uniformly accelerated along the straight line
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The magnitude of the acceleration along the straight line
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The motion is simple harmonic about ${x=\sqrt{12} m}$
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The maximum displacement from the fixed point is 4 cm

Explanation

${ v^{ 2 }=108-9x^{ 2 } }\\ \Rightarrow v=\sqrt { 108-9x^{ 2 } } =\sqrt { 9\times 12-9x^{ 2 } } =3\sqrt { 12-x^{ 2 } }$
comparing it with the velocity in SHM

$v=\omega \sqrt { { A }^{ 2 }-{ x }^{ 2 } }$ we have

$\omega=3s^{-1}$ and

$A=\sqrt{12}m$
the given equation represents a SHM about

$x=0$ .

$\Rightarrow$ C is incorrect.
In SHM
(I) acceleration is not uniform

$\Rightarrow$ A is incorrect.
(II) acceleration is along a straight line.

$\Rightarrow$ B is correct.
In the given SHM maximum displacement from the mean position is

$A=\sqrt{12}m$

$\Rightarrow$ D is incorrect.

A particle executes SHM with time period

$T$ and amplitude

$A$ .The maximum possible average velocity in time

$\displaystyle{\dfrac{T}{4}}$ is

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$\displaystyle\dfrac{2A}{T}$
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$\displaystyle\dfrac{4A}{T}$
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$\displaystyle\dfrac{8A}{T}$
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$4\sqrt{2} \dfrac{A}{T}$

Explanation

Maximum average velocity magnitude for

$\dfrac{T}{4}$ time will occur in the time from

$\dfrac{-T}{8}$ to

$0$ to

$\dfrac{+T}{8}$ ,
we can Integrate

$A \omega \cos{\omega t} {dt}$ within the limits from

$\dfrac{-T}{8}$ to

$\dfrac{+T}{8}$ , and divide the answer by

$\dfrac{T}{4}$
here

$\omega=\dfrac{2\pi}{T}$ , that gives, the maximum average velocity

$4\sqrt{2} \dfrac{A}{T}$ .

A particle moves along the x-axis according to:

${x=A\left [ 1+sin \omega t \right ]}$ . What distance does it travel between t=0 and

${t=2.5\pi/\omega}$ ?

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4A
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6A
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5A
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None

Explanation

In the given SHM, particle starts from the extreme position. and the angular frequency of the given SHM is

$\omega$ .
So the time period of the given SHM will be

$\dfrac{2\pi}{\omega}s$
We need to calculate the distance in

$t=\dfrac{2.5\pi}{\omega}s=\dfrac{2\pi}{\omega}s+\dfrac{0.5\pi}{\omega}s$
or

$t=\left(T+\dfrac { T }{ 4 }\right) s$
in one time period

$(T)$ it will come back to its initial position so the distance traveled in time

$T s$ is

$4A$ and after that it will reach to mean position in time

$\frac{T}{4}s$ , so the distance traveled in this time

$\dfrac{T}{4}s$ will be

$A$ . So the total distance traveled in time

$t=\left(T+\dfrac { T }{ 4 }\right) s$ will be equal to

$4A+A=5A$

A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude A.Its speed as it passes through the equilibrium position is V.If extended 2A and released, the speed of the mass passing through the equilibrium position will be

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2V
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4V
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V/2
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V/4

Explanation

At equilibrium position the speed of a pendulum is equal to

$A\omega$ . If we double the amplitude, the speed will also double. so the correct option is 'A'

How long will it be moving until it stops for the first time?

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$\displaystyle t=2\frac{\pi}{\omega}$
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$\displaystyle t=3\frac{\pi}{\omega}$
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$\displaystyle t=4\frac{\pi}{\omega}$
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$\displaystyle t=\frac{\pi}{\omega}$

Explanation

The equation of force is similar to that in case of a SHM.
Time period of a SHM is given by,

$t = (2\times \pi )/\omega$
But we know that the particle undergoing SHM comes to rest twice in one cycle.
Time when the particle first comes to rest

$t = \pi/\omega$

A particle is moving in a circular path with continuously increasing speed. Its motion is

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periodic but not oscillatory
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periodic but not SHM
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oscillatory
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none of these

Which of the following quantities is always negative in SHM?

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$\vec{F}.\vec{a}$
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$\vec{F}.\vec{r}$
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$\vec{v}.\vec{r}$
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$\vec{a}.\vec{r}$

Explanation

Since force and acceleration is always in opposite direction of displacement. so the dot product of displacement $\vec r$ with $\vec F$ and $\vec a$ is always smaller than zero. $\Rightarrow$ B and D are correct
Force and acceleration are always in same direction so dot product of $\vec F$ and $\vec a$ is always greater then zero. $\Rightarrow$ A is incorrect
Velocity and displacement are not always in the same direction so dot product of

$\vec v$ and $\vec r$ is sometimes greater then zero and sometimes less then zero $\Rightarrow$ C is incorrect

Which of the following quantities are always zero in SHM?

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$\vec{F}\times \vec{a}$
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$\vec{v}\times \vec{r}$
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$\vec{a}\times \vec{r}$
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$\vec{F}\times \vec{r}$

Explanation

In SHM

$\vec { F }$ ,

$\vec { r }$ ,

$\vec { v }$ and

$\vec { a }$ are always along the same straight line either in the same direction or in opposite directions, so the angle between any of the pairs is either

${ 0 }^{ \circ }$ or

${ 180 }^{ \circ }$ and

$\sin { { 0 }^{ \circ } } =\sin { { 180 }^{ \circ } } =0$ . So cross product of any of the pair will also be equal to zero. Hence all options are correct.

The phase angle between the projections of uniform circular motion on two mutually perpendicular diameter is

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$\pi$
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$3\pi/4$
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$\pi/2$
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zero

Explanation

when one projection will be in mean position other will be in the extreme position. so the phase difference will be

$\dfrac {\pi}{2}$

For a particle executing SHM having amplitude 'a' the speed of the particle is one half of its maximum speed when its displacement from the mean position is

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$a/2$
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a
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$a\frac{\sqrt{3}}{2}$
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2a

Explanation

$x=A\sin { \omega t } \\ \Rightarrow \dfrac { dx }{ dt } =A\omega \cos { \omega t } \\ \Rightarrow v=A\omega \cos { \omega t }$

$v=\dfrac { v_{ max } }{ 2 } =\dfrac { A\omega }{ 2 }$

$\Rightarrow \dfrac { A\omega }{ 2 } =A\omega \cos { \omega t } \\ \Rightarrow \cos { \omega t } =\dfrac { 1 }{ 2 } \\ \Rightarrow \sin { \omega t } =\sqrt { 1-{ \left( \dfrac { 1 }{ 2 } \right) }^{ 2 } } =\sqrt { 1-{ \dfrac { 1 }{ 4 } } } =\dfrac { \sqrt { 3 } }{ 2 } \\ \Rightarrow x=a\sin { \omega t } =\dfrac { a\sqrt { 3 } }{ 2 }$

What is the number of degrees of freedom of an oscillating simple pendulum?

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more than three
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3
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2
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1

The circular motion of a particle with constant speed is:

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periodic but not SHM
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SHM but not periodic
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periodic and SHM
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nethier periodic nor SHM

A person wearing a wrist watch that keeps correct time at the equator goes to N-pole. His watch will

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Keep correct time
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gain time
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loose time
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cannot say

The dimensional formula for amplitude of SHM is

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$MLT$
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$M^{\circ}L^{\circ}T^{\circ}$
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$M^{\circ}LT^{\circ}$
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$MLT^{\circ}$

Explanation

Amplitude is the maximum displacement from mean level. Thus it has dimensions of Length or

$M^0LT^0$

A mass m is suspended from a spring of force constant

$k.$ The angular frequency of oscillation of the spring will be

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$\dfrac{k}{m}$
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$\sqrt{\dfrac{m}{k}}$
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$\dfrac{m}{k}$
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$\sqrt{\dfrac{k}{m}}$

Explanation

Referring Formula 1, we get

$\omega =\sqrt{\frac{k}{m}}$ . Hence, option D is correct.

A body of mass 100 gm is suspended from a spring of force constant 50 N/m. The maximum acceleration produced in the spring is:

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g/2
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g
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g/3
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g/4

Explanation

To calculate maximum acceleration consider spring at the equilibrium position:

$F=ma$
where

$F$ is force on spring which will be equal to

$mg$ only (as gravity is the only force present.
Therefore,

$ma=mg$ or

$a=g$ .

In SHM, which of the following quantities does not vary as per nature of the sine curve?

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acceleration
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time period
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displacement
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velocity

Explanation

$x=a\sin\omega t$ .
Hence on differentiating, we get

$v=a\omega \cos\omega t$ and

$a=-a\omega^2 \sin\omega t$
We can hence see that

$x,\ v$ and

$a$ are all dependent on sine curve, but time is not.

Which of the following characteristics must remain constant for undamped oscillations of the particle?

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acceleration
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phase
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amplitude
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velocity

The displacement of a particle executing SHM at any instant t is

$x=0.01 \sin{100(t+0.05)}$ then its time period will be

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0.06 s
0%
0.2 s
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0.1 s
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0.02 s

Explanation

$x=0.01\sin(100t+5)$
We have

$x=A\sin(\omega t + \theta)$

On comparing we get
Thus, here

$\omega=100$
Thus,

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{100}=0.06\ \text{seconds}.$

A particle is moving on a circle with uniform speed. Its motion is

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a periodic motion
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periodic and SHM
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periodic but not SHM
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none of these

Two particles P and Q describe SHM of same amplitude a and frequency v along the same straight line. The maximum distance between the two particles is

$a\sqrt{2}$ . The initial phase difference between the particle is

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$\pi/3$
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$\pi/2$
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$\pi/6$
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zero

Explanation

Let initial phase difference be

$\theta$

So, distance between two particales is

$d=asin(\omega t+\theta)-asin(\omega t)=2asin(\theta/2)cos(\omega t+\theta/2)$

As maximum value of

$d$ is

$a\sqrt 2$ , so

$2asin(\theta/2)=a\sqrt 2\Rightarrow \theta=\pi/2$

The phase difference between the velocity and displacement of a particle executing SHM is

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$\pi/2$ radian
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$\pi$ radian
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$2\pi$ radian
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zero

Explanation

$x=Asin(\omega t +\theta)$

$v=dx/dt=A\omega cos(\omega t + \theta)=-A\omega sin (\omega t + \theta + \pi/2)$ since

$cos(\pi/2+x)=-sin x$
Thus, phase difference is of

$\pi/2$

If at any instant of time the displacement of a harmonic oscillator is

$0.02\ \text{m}$ and its acceleration is

$2\ \text{ms}^{-2}$ , its angular frequency at that instant will be

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$0.1\ \text{rads}^{-1}$
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$1\ \text{rads}^{-1}$
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$10\ \text{rads}^{-1}$
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$100\ \text{rads}^{-1}$

Explanation

At any instant, neglecting the direction, we have,

$a=\omega ^{2}x$ .
Substituting the given values, we get

$\omega=10 rad/s$ . Hence, option C is correct.

The equation of displacement of a particle executing SHM is

$x=0.40 \cos(2000 t+18)$ . The frequency of the particle is

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$10^3\ \text{Hz}$
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$20\ \text{Hz}$
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$2\times 10^3\ \text{Hz}$
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$\displaystyle \dfrac{10^3}{\pi}\ \text{Hz}$

Explanation

$\omega = 2000$ here
Thus,

$f=\dfrac{\omega}{2\pi}=\dfrac{1000}{\pi}\ \text{Hertz}$

Identical springs of spring constant

$K$ are connected in series and parallel combinations. A mass

$m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

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1:4
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1:2
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4:1
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2:1

Explanation

From Formula 1. in series combination, ,

$k_{eq}=\frac{K\times K}{K+K}=\dfrac{K}{2}$ .

$\therefore f_{1}=\dfrac{1}{2\pi }\sqrt{\dfrac{K}{2m}}$ ...(1)
For parallel combination,

$k_{eq}=2K$

$\therefore f_{2}=\dfrac{1}{2\pi }\sqrt{\dfrac{2K}{m}}$ ...(2)
Taking ratio, we get answer as

$1:4$ Option A is correct.

The following figure shows three identical springs

$A, B, C$ . When a 4 kg weight is hung on

$A$ , it descends by

$1$ cm. When a

$6$ kg weight is hung on

$C$ , it will descend by

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1.5 cm
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3.0 cm
0%
4.5 cm
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6.0 cm

Explanation

The spring constant can be found out by

$k=\dfrac{F}{x}=\dfrac{4\times 10}{0.01}=4000N/m$
Now, when two springs are connected in series, their equivalent constant is

$\dfrac{1}{k_{eq}}=\dfrac{1}{4000}+\dfrac{1}{4000}$

$k_{eq}=2000N/m$
Therefore, mass of 6 kg will descend by

$x=\dfrac{6\times 10}{2000}=0.03m=3cm$
Option B is thus correct.

In SHM, select the wrong statement, where

${F}$ is the force,

${a}$ is the acceleration and

${v}$ is the velocity of the particle in SHM.

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$\overset{\rightarrow}{F}\times \overset{\rightarrow}{v}$ is a null vector
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$|\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}|=0$ (always)
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$\overset{\rightarrow}{F}\times \overset{\rightarrow}{a}< {0}$
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$\overset{\rightarrow}{a}.\overset{\rightarrow}{x}<0$

Explanation

The velocity, acceleration and thus the force are along the direction of motion in the SHM.

So the cross product of Force with velocity or acceleration is always zero.

Thus

$A$ and

$B$ are correct.

Also, acceleration is always in opposite direction to the displacement vector. So dot-product of acceleration and displacement is always negative. Thus

$D$ is also correct.

$C$ is the wrong statement as the cross product of force and acceleration is strictly zero, because,

$F=ma\Rightarrow ma\times a=0$ .

A particle of mass

$M$ is executing oscillations about the origin on the x-axis. Its potential energy is

$|U|=K|x^2|$ where

$K$ is a positive constant. If the amplitude of oscillations is

$a$ , then its period

$T$ is

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proportional to $1/\sqrt{a}$
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independent of $a$
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proportional to $\sqrt{a}$
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proportional to $a^{1/2}$

Explanation

The restoring force is

$-m\omega^2x=-dU/dx=-2Kx\Rightarrow \omega=\sqrt{2K/m}$ where

$\omega$ is the angular frequency of SHM.

Time period

$T=2\pi/\omega=2\pi\sqrt{m/2K}$

So Time period is independent of amplitude

$a$ .

The equation of S.H.M. of a particle is

$a+4\pi^{2}x=0$ where

$a$ is the instantaneous linear acceleration at displacement

$x$ . The frequency of motion is

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$1\space Hz$
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$4\pi\space Hz$
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$\frac{1}{4}\space Hz$
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$4\space Hz$

Explanation

Comparing

$a+\omega^{2}x=0\ with\ a+4\pi^2x=0$

$\omega^{2}= 4\pi^{2} \Rightarrow \omega=2\pi , 2\pi n=2\pi , n=1\ Hz$

Which of the following quantities is non-zero at the mean position for a particle executing SHM?

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force
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acceleration
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velocity
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displacement

Explanation

At mean position, displacement,

$x=0$ . Since for SHM, acceleration is directly proportional to

$x$ , so acceleration is also 0.
Hence

$F=ma$ is also zero.
But velocity is not zero (it is zero at extreme positions)

A particle is executing SHM along a straight line

$8\ \text{cm}$ long. While passing through mean position its velocity is

$16\ \text{cms}^{-1}$ . Its time period will be:

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$0.0157\ \text{s}$
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$0.157\ \text{s}$
0%
$1.57\ \text{s}$
0%
$15.7\ \text{s}$

Explanation

8 cm long line means amplitude is 4 cm (both sides of mean)
Velocity at mean position is the maximum velocity given by

$A\omega$
Thus

$16=4\omega$ or

$\omega = 4$
Thus,

$T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{4}=\dfrac{\pi}{2}=1.57\ \text{s}$

The time period of a particle executing SHM is

$\displaystyle \frac{2\pi}{\omega}$ and its velocity at a distance

$b$ from mean position is

$\sqrt{3}b\omega$ . Its amplitude is

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b
0%
2b
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3b
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4b

Explanation

$x=A\sin\omega t$

$v=A\omega \cos\omega t$
Thus, when

$x=b,$ we have

$\sin\omega t = \dfrac{b}{A}$

and hence,

$\cos\omega t=\sqrt{1-\left(\dfrac{b}{A}\right)^2}$

Thus,

$\sqrt 3 b\omega = A\omega \sqrt{1-\left(\dfrac{b}{A}\right)^2}$

or

$3b^2 = A^2\left(1-\dfrac{b^2}{A^2}\right)$

or

$3b^2=A^2-b^2$
Thus,

$A^2=4b^2$ or

$A=2b$

Restoring force in the SHM is

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conservative
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nonconservative
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frictional
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centripetal

Explanation

A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Restoring force

$F=kx,$ is such kind of force and is conservative.

Due to some force

$F_1$ , a body oscillates with period 4/5 second and due to other force

$F_2$ it oscillates with the period of 3/5 sec. If both forces act simultaneously new period will be

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0.72 s
0%
0.64 s
0%
0.48 s
0%
0.36 s

Explanation

for first case,

$\omega_1^2=\dfrac{F_1}{mx}$

for second case,

$\omega_2^2=\dfrac{F_2}{mx}$

When both forces applied,

$\omega_3^2=\dfrac{F_1+F_2}{mx}$

$\Rightarrow \omega_3^2= \omega_1^2+ \omega_2^2$

$\Rightarrow \omega_3^2/4\pi^2= \omega_1^2/4\pi^2+ \omega_2^2/4\pi^2$

$\dfrac{1}{T_3^2}=\dfrac{1}{T_1^2}+\dfrac{1}{T_2^2}$

$T_1=4/5=0.8s;T_2=3/5=0.6s$

$\Rightarrow T_3=12/25=0.48\ sec$

Statement 1: The graph between velocity and displacement for a harmonic oscillator is an ellipse.

Statement 2: Velocity does not change uniformly with displacement in harmonic motion.

0%
Statement 1 is false, Statement 2 is true
0%
Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation for Statement 1
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Statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation for Statement 1
0%
Statement 1 is true, Statement 2 is false

Explanation

$x=Asin\omega t$ or

$sin\omega t=x/A$

$v=dx/dt=A\omega cos\omega t$ or

$cos\omega t=v/A\omega$
Thus, using

$sin^2+cos^2=1$ ,

$x^2/A^2+x^2/A^2\omega^2=1$ , which gives the equation of ellipse.
Thus statement 1 is correct. Also, statement 2 is correct but it doesn't explain the reason.

For a particle undergoing simple harmonic motion, the velocity is plotted against displacement. The curve will be

0%
a straight line
0%
a parabola
0%
a circle
0%
an ellipse

Explanation

$v=\omega \sqrt{A^{2}-x^{2}}$

$\therefore$

$\displaystyle \dfrac{v^{2}}{\omega ^{2}}+\dfrac{x^{2}}{\left ( 1 \right )^{2}}=A^{2}$
Hence

$v-x$ graph is an ellipse.

A particle executes simple harmonic motion with frequency 2.5 Hz and amplitude 2 m. The speed of the particle 0.3 s after crossing the equilibrium position is:

0%
Zero
0%
$2\pi$ m/s
0%
$4\pi$ m/s
0%
$\pi$ m/s

Explanation

$\displaystyle T=\frac{1}{F}=\frac{1}{2.5}=0.4$ s

$\displaystyle t=0.3s=\frac{3T}{4}$
At the given time, particle win be in its extreme position if at t=0 it crosses the mean position.

A simple harmonic oscillator has an acceleration of

$1.25\ m/s^{2}$ at

$5\ cm$ from the equilibrium. Its period of oscillation is:

0%
$\displaystyle \dfrac{4\pi }{5}$ s
0%
$\displaystyle \dfrac{5\pi }{2}$ s
0%
$\displaystyle \dfrac{2\pi }{5}$ s
0%
$\displaystyle \dfrac{2\pi }{25}$ s

Explanation

$\displaystyle T=2\pi \sqrt{\left | \frac{x}{a} \right |}=2\pi \sqrt{\frac{5\times 10^{-2}}{1.25}}=\frac{2\pi }{5}$ sec

A mass is suspended separately by two springs of spring constant

$k_{1}$ and

$k_{2}$ in successive order. The time periods of oscillations in the two cases are

$T_{1}$ and

$T_{2}$ respectively. If the same mass be suspended by connecting the two springs in parallel

$($ as shown in figure

$)$ then the time period of oscillations is

$T.$ The correct relation is

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$T^{2}=T_{1}^{2}+T_{2}^{2}$
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$T^{-2}=T_{1}^{-2}+T_{2}^{-2}$
0%
$T^{-1}=T_{1}^{-1}+T_{2}^{-1}$
0%
$T=T_{1}+T_{2}$

Explanation

$\displaystyle T=2\pi \sqrt{\frac{M}{k}}$

$\therefore$

$\displaystyle k=\frac{4\pi ^{2}m}{T^{2}}$
In the given situation

$k=k_{1}+k_{2}$

$\therefore$

$\displaystyle \frac{4\pi ^{2}m}{T^{2}}=\frac{4\pi ^{2}m}{T_{1}^{2}}+\frac{4\pi ^{2}m}{T_{2}^{2}}$

$\therefore$

$T^{-2}=T_{1}^{-2}+T_{2}^{-2}$

A particle moves in a circular path with a uniform speed. Its motion is:

0%
Periodic
0%
Oscillatory
0%
Simple harmonic
0%
Angular simple harmonic

Explanation

$\textbf{Correct option: Option (A).}$

$\textbf{Explanation:}$

The motion of particle moving in a circular path with uniform speed is periodic and is not oscillatory because it’s not to and fro. Hence it can also not be SHM and angular SHM.

$\textbf{Thus, option (A) is only correct.}$

Statement 1: In simple harmonic motion, the motion is to and fro and periodic.

Statement 2: Velocity of the particle

$(v) = \omega\sqrt{k^2 - x^2}$ (where

$x$ is the displacement).

0%
Statement 1 is false, Statement 2 is true
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Statement 1 is true, Statement 2 is true; Statement 2 is the correct explanation for Statement 1
0%
Statement 1 is true, Statement 2 is true; Statement 2 is not the correct explanation for Statement 1
0%
Statement 1 is true, Statement 2 is false

Explanation

$S.H.M$ is to and fro motion of an object and it is periodic.

$v=\omega \sqrt{k^2 - x^2}$
If

$x = 0$ ,

$v$ has maximum value. At

$x = k$ ,

$v$ has minimum velocity. Similarly, when

$x = -k$ ,

$v$ has zero value, all these indicate to and fro movement.

A particle of mass 2 kg moves in simple harmonic motion and its potential energy U varies with position x as shown. The period of oscillation of the particle is:

0%
$\displaystyle \frac{2\pi }{5}$ s
0%
$\displaystyle \frac{2\sqrt{2}\pi }{5}$ s
0%
$\displaystyle \frac{\sqrt{2}\pi }{5}$ s
0%
$\displaystyle \frac{4\pi }{5}$ s

Explanation

$\displaystyle \frac{1}{2}kA^{2}=1.0$

$\displaystyle \frac{1}{2}k\left ( 0.4 \right )^{2}=1.0$
or

$k=12.5$ or

$\displaystyle \frac{25}{2}$ N/m

$\displaystyle T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{2}{\frac{25}{2}}}=\frac{4\pi }{5}$ s

Select the correct statements from the following

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A simple harmonic motion is necessarily periodic
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A simple harmonic motion may be oscillatory
0%
An oscillatory motion is necessarily periodic
0%
A periodic motion is necessarily oscillatory

Explanation

$\textbf{Correct option: Option (A).}$

$\textbf{Explanation:}$

(A) A simple harmonic motion is always repeated at a regular interval and hence it is necessarily periodic.

(B) An SHM is a to and fro motion and hence it is necessarily oscillatory and hence this statement is incorrect.

(D) A periodic motion may or may not be oscillatory. So this statement is also false.

$\textbf{Thus, only option (A) is correct.}$

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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