MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 2

The equation of motion of a particle is

$x = a \cos (a t)^{2}$ . The motion is

0%
periodic but not oscillatory.
0%
Periodic and oscillatory.
0%
oscillatory but not periodic
0%
neither periodic nor oscillatory

Explanation

$x = a \cos (a t)^{2}$ here,

$x$ is a cosine.

This results in oscillatory motion.

$f(t)= f(x)+ f(t)= f(x+t)$

$x(t)= a \cos \alpha {t} ^{2}$

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

$a\cos(\ alpha {t}^{2} )+ (\alpha {T}^{2}) +(2\alpha t T)$

which is not equal to

$x(t)$ this shows this is not periodic.

The displacement of a particle varies with time, according to the equation

$y = a \sin \omega t + b \cos \omega t$

0%
The motion is oscillatory but not SHM.
0%
The motion is SHM. with amplitude (a + b)
0%
The motion is SHM. with amplitude $(a^{2} + b^{2})$
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The motion is SHM. with amplitude $\sqrt{a^{2} + b^{2}}$

Explanation

(d) :

$y = a \sin \omega t + b \cos \omega t$

The amplitude of motion is given by

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

Hence, verifies the option (d).

If a simple pendulum is taken to place where

$g$ decreases by

$2\%$ then the time period

0%
Decreases by $1\%$
0%
Increases by $2\%$
0%
Increase by $2\%$
0%
Increases by $1\%$

Explanation

Time period for simple pendulam,

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

$\Rightarrow T\propto \dfrac{1}{\sqrt{g}}$

$\Rightarrow \dfrac{\Delta T}{T}\times 100=-\dfrac{1}{2}\left(\dfrac{\Delta g}{g}\right)\times 100=-\dfrac{1}{2}(-2\%)=1\%$

A simple pendulum is vibrating in an evacuated chamber, it will oscillate with

0%
Increasing amplitude
0%
Constant amplitude
0%
Decreasing amplitude
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First $(c)$ then $(a)$

The velocity of a particle performing simple harmonic motion, when it passes through its mean position is

0%
Infinity
0%
Zero
0%
Minimum
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Maximum

Explanation

Equation of SHM particle

$Y=a\sin \omega t\\V=a\omega \sin\omega t$

$V_{max}=a\omega$

The length of simple pendulum is increased by

$1\%$ . Its period will

0%
Increase by $1\%$
0%
Increase by $0.5\%$
0%
Decrease by $0.5\%$
0%
Increase by $2\%$

Explanation

$T\propto \sqrt {l}$

$\Rightarrow \dfrac {\Delta T}{T}=\dfrac {1}{2}\dfrac {\Delta l}{l}=\dfrac {1}{2}\times 1\% =0.5\%$

The velocity of a particle that executes S.H.M. is at its mean position 0.866 m/s.Velocity at a displacement half of its amplitude from mean position is

0%
1 m $s^{-1}$
0%
1.414m $s^{-1}$
0%
0.5m $s^{-1}$
0%
0.75m $s^{-1}$

Explanation

$v_{max}=A\omega = 0.866\dfrac{m}{s}$
At $x=\dfrac{A}{2} ; v=\omega \sqrt{A^{2}-x^{2}}$
$v=\dfrac{\sqrt{3}}{2}\omega A=\dfrac{\sqrt{3}}{2}v_{max}$
$v=\dfrac{\sqrt{3}}{2}\times 0.866=0.75\dfrac{m}{s}$

The equation for the displacement of a particle executing SHM is y = {5 sin 2

$\pi$ t }cm. Then the velocity at 3 cm from the mean position is

0%
2 $\pi$ cm/s
0%
3 $\pi$ cm/s
0%
4 $\pi$ cm/s
0%
8 $\pi$ cm/s

Explanation

$\omega =2\pi \ {rad}/{sec}$

$A=5cm$

$x=3cm$

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

$=2\pi \times \sqrt{25-9}=8\pi \ {cm}/{s}$

The equation of the displacement of two particles making SHM are represented by

$y_{1}$ = a sin

$\left ( \omega t +\phi \right )$ &

$y_{2}$ = a cos

$\left ( \omega t \right )$ .

The phase difference of the velocities of the two particles is :

0%
$\frac{\pi }{2} +\phi$
0%
$- \phi$
0%
$\phi$
0%
$\phi -\frac{\pi }{2}$

Explanation

$y_{1}=a sin (\omega t+\phi )$

$y_{2}=a cos (\omega t)$

$y_{1}=a \omega cos (\omega t+\phi)=a \omega sin (\omega t+\phi +{\pi}/_{2})$

$y_{2}=-a \omega sin (\omega t)= a \omega sin (\pi + \omega t)$
therefore phase difference

$= \omega t+\phi +{\pi}/_{2}-(\pi+\omega t)$

$=\phi -{\pi}/_{2}$

$T _{1}$ ,

$T _{2}$ are time periods of oscillation of a block individually suspended to spring of force constants

$K _{1}$ ,

$K _{2}$ respectively. If same block is suspended to parallel combination of same two springs, its time period is

0%
$T_{1}+T_{2}$
0%
$\dfrac{T_{1}+T_{2}}{2}$
0%
$\dfrac{T_{1}T_{2}}{T_{1}+T_{2}}$
0%
$\dfrac{T_{1}T_{2}}{\sqrt{{T_{1}}^{2}+{T_{2}}^{2}}}$

Explanation

for a parallel combination $k=k_{1}+k_{2}$
$T=2\pi \sqrt{\dfrac{m}{k}}$

$=2\pi \sqrt{\dfrac{m}{(k_{1}+k_{2})}}$

$T_{1}=2\pi \sqrt{\dfrac{m}{k_{1}}}$

$\Rightarrow k_{1}=(2\pi )^{2}\dfrac{m}{T_{1}^{2}}$

$k_{2}=(2\pi )^{2}\dfrac{m}{T_{2}^{2}}$

$k_{1}+k_{2}=(2\pi )^{2}(\dfrac{m}{T_{1}^{2}}+\dfrac{m}{T_{2}^{2}})$

$\therefore \dfrac{T_{1}T_{2}}{\sqrt{T_{1}+T_{2}}}=T$

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is

$15 cm/sec$ and the period is

$628$ milliseconds. The amplitude of the motion in centimeters is:

0%
$3$
0%
$2$
0%
$1.5$
0%
$1.0$

Explanation

$v_{max}=15\ cm /s ; T=0.628\ s=\dfrac{2\pi}{10}s$
$\omega =\dfrac{2\pi }{T}=10\ rad/s$
$v_{max}=A\omega$
$A=1.5\ cm$

Assertion (A): The phase difference between displacement and velocity in SHM is

$90^{\circ}$
Reason (R): The displacement is represented by y=A sin

$\omega$ t and Velocity by V=A

$\omega$ cos

$\omega$ t.

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true and R is not the correct explanation of A
0%
A is true and R is false
0%
A is false and R is true

Explanation

At mean displacement is Zero but velocity is maximum At extreme position displacement is maximum but velocity is Zero.
Thus displacement can be represented by

$y$ =

$A \ sin \ wt$

$V=\dfrac{dy}{dt}$

$=Awcoswt= Aw sin (wt+\dfrac{\pi}{2})$
Thus phase difference is

$\dfrac{\pi}{2}$

If the displacement

$x$ and velocity

$v$ of a particle executing S.H.M are related through the expression

$4v^{2}=25-x^{2}$ , then its maximum displacement in meters is

0%
1
0%
2
0%
5
0%
6

Explanation

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

$4v^{2}=5^{2} -x^{2} \implies v=\frac{1}{2} \sqrt{(5^{2}-x^{2})}$
Comparing the equation with the standard equation we get

$A =$ 5m

The equation of motion of a particle in S.H.M. is given by

$a = -B \times x$ ,where a is the acceleration, B is a constant and x is displacement. The period of oscillation of the particle is:

0%
2 $\pi \sqrt{B}$
0%
2 $\pi / B$
0%
2 $\pi/\sqrt{B}$
0%
2 $\pi B$

Explanation

$a=-Bx$
so,

$B=\omega^{2}$

$(\therefore a=-\omega^{2}x)$

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

$=\dfrac{2\pi}{\sqrt{B}}$

$so, time\ period \ of\ oscillations=\dfrac{2\pi}{\sqrt{B}}$

The amplitude of oscillation of a particle is 0.05 m.If its period is 1.57s. Then the velocity at the mean position is

0%
0.1 m/s
0%
0.2 m/s
0%
0.3 m/s
0%
0.5 m/s

Explanation

$A=0.05m$

$T=\dfrac{\pi }{2}sec ; w=4\ { rad}/{sec}$

$v_{max}=wA =0.2 \ {m}/{s}$

The time period of oscillation of a particle that executes S.H.M. is

$1.2 sec$ . The time starting from extreme position, its velocity will be half of its velocity at mean position is

0%
$0.1$ s
0%
$0.2$ s
0%
$0.4$ s
0%
$0.6$ s

Explanation

Time is zero at extreme position

$T=1.2 sec$

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

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

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

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

$x=A$ cos

$\omega t$

$\dfrac{\sqrt{3}}{2}A=A$ cos

$\omega t$

$\omega t=\dfrac{\pi }{6}$

$t=\dfrac{\pi }{6\times \omega }$

$t=\dfrac{\not{\pi} }{6}\times \dfrac{1.2}{{2\not \pi }}=0.1 sec$

A simple harmonic oscillation is represented by the equation

$y = 0.4 \sin(440t/7+0.61)$ where

$y$ and

$t$ are in meters and seconds respectively, the value of time period is :

0%
$0.1\ s$
0%
$0.2\ s$
0%
$0.3\ s$
0%
$0.4\ s$

Explanation

$\omega =\dfrac{440}{7}$

$\dfrac{2\pi }{T}=\dfrac{440}{7} ; T=\dfrac{14\pi }{440}$

$T=0.1 sec$

The velocities of a particle performing linear SHM are

$0.13$ m/s and

$0.12$ m/s when it is at

$0.12m$ and

$0.13$ m from mean position respectively, the period is:

0%
$\pi$ s
0%
2 $\pi$ s
0%
3 $\pi$ s
0%
4 $\pi$ s

Explanation

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

$V_{1}=\omega\sqrt{A^{2}-x^{2}_{1}}$

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

$\dfrac{0.13}{0.12}$ =

$\dfrac{\sqrt{A^{2}-(0.12)^{2}}}{\sqrt{A^{2}-(0.13)^{2}}}$

$\dfrac{0.13^2}{0.12^2}$ =

$\dfrac{{A^{2}-(0.12)^{2}}}{{A^{2}-(0.13)^{2}}}$

Cross-multiplying,

$\left ((0.13)^{2}-(0.12)^{2}\right) A^2 =(0.13)^{4}-(0.12)^{4}$

$A^{2}=(0.13)^{2}+(0.12)^{2}$

$A=\sqrt{(0.13)^{2}+(0.12)^{2}}$

Using,

$x$ =

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

$0.13=\omega \sqrt{(0.13)^{2}+(0.12)^{2}-(0.12)^{2}}$

$\therefore \omega =1$

$T=2\pi /\omega$

$T=2\pi \sec$

A body executes S.H.M. under the action of a force

$F$

$_{1}$ with a time period

$4/5$ seconds. If the force is changed to

$F$

$_{2}$ , it executes S.H.M. with a time period

$3/5$ seconds. If both the forces

$F$

$_{1}$ and

$F$

$_{2}$ act simultaneously in the same direction on the body, then its time period in seconds is:

0%
$12/25$
0%
$24/25$
0%
$25/24$
0%
$25/12$

Explanation

$F_{1}=m\omega^{2}x$

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

$F=ma=$

$4m\pi^{2} x \dfrac{1}{T^{2}}$

$\therefore F_{1}=\dfrac{4m\pi^{2}}{T_{1}^{2}}x$

$F_{2}=\dfrac{4m \pi^{2}}{T^{2}_{2}}x$

$F_{1}+F_{2}=\dfrac{ 4m \pi^{2}x}{T^{2}} + \dfrac{4m \pi^{2}x}{T^{2}_{2}}$

$\dfrac{1}{T^{2}}=\dfrac{1}{T^{2}_{1}}+\dfrac{1}{T^{2}_{2}}$

$\therefore T=\dfrac{12}{25}$

A particle executing SHM has velocities

$20$ cm/s and

$16$ cm/s at displacements

$4$ cm and

$5$ cm from its mean position respectively. Its time period is

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

Explanation

$v_{1}=\omega _{1}\sqrt{A^{2}-x_{1}^{2}}$

$v_{2}=\omega _{1}\sqrt{A^{2}-x_{2}^{2}}$

$\dfrac{v_{1}}{v_{2}}=\dfrac{\sqrt{A^{2}-x{_{1}}^{2}}}{\sqrt{A^{2}-x{_{2}}^{2}}} ; v{_{1}}^{2}A^{2}-v{_{1}}^{2}x{_{2}}^{2}=v{_{2}}^{2}A^{2}-v{_{2}}^{2}x{_{1}}^{2}$

$A^{2}=\dfrac{v{_{1}}^{2}x{_{2}}^{2}-v{_{2}}^{2}x{_{1}}^{2}}{v{_{1}}^{2}-v{_{2}}^{2}}=\dfrac{400\times 25-16\times 256}{400-256}$

$A^{2}=41$
Substituting

$A^{2}=41$ we get

$\omega ^{2}=16$

$\omega =\dfrac{2 \pi}{T}=4 ; T=\dfrac{2\pi }{4}=\dfrac{\pi }{2}sec$

The velocity of a particle in SHM at the instant when it is

$0.6$ cm away from the mean position is

$4$ cm /sec. If the amplitude of vibration is

$1$ cm then its velocity at the instant when it is

$0.8$ cm away from the mean position is:

0%
$2.25$ cm/s
0%
$2.5$ cm/s
0%
$3.0$ cm/s
0%
$3.5$ cm/s

Explanation

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

$V_{1}=\omega \sqrt{A^{2}-x^{2}}$

$V_{1}=\omega \sqrt{A^{2}-(0.6)^{2}}$

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

$V_{2}=\omega \sqrt{A^{2}-(0.8)^{2}}$

$\dfrac{V_{1}}{V_{2}}=\dfrac{\sqrt{(1^{2}-(0.6)^{2}}}{\sqrt{(1^{2}-(0.8)^{2}}}$

$\dfrac{4}{V_{2}}=\dfrac{0.8}{0.6}$

$V_{2}=3 \ cm / sec$

Two S.H.M.'s are represented by the relations

$x=4sin(80t+\pi /2)$ and

$y=2cos(60t+\pi/3)$ .

The ratio of their time periods is

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2:1
0%
1:2
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4:3
0%
3:4

Explanation

$x=4 sin(80t +\dfrac{\pi }{2})$
$\omega _{1}=80 ;\dfrac{2\pi}{T_{1}}=80 ; T_{1}=\dfrac{\pi }{40}$ sec
$y=2 cos(60t + \dfrac{\pi }{3})$
$\omega _{2}=60 ; \dfrac{2\pi }{T_{2}}=60 ; T_{2}=\dfrac{\pi }{30}sec$
$\dfrac{T_{1}}{T_{2}}=\dfrac{3}{4}$

The displacement of a particle executing S.H.M. is given by

$y = 10 \sin (6t+\pi/3)$ in metre and time in seconds. The initial displacement and velocity of the particle are respectively :

0%
$25\ m$ and $5\sqrt{3}\ m/s$
0%
$15\ m$ and $5\sqrt{3}\ m/s$
0%
$15\sqrt{3}\ m$ and $30\ m/s$
0%
$10\sqrt{3}/2\ m$ and $30\ m/s$

Explanation

Initially

$t=0 ;y=10 sin(\dfrac{\pi }{3})$

$=10\times \dfrac{\sqrt{3}}{2}=10\dfrac{\sqrt{3}}{2}m$

$v=\dfrac{dy}{dt}=60 cos(6t + \dfrac{\pi }{3})$

$t=0 \Rightarrow 60\times cos \dfrac{\pi }{3}=30{m/}{s}$

A body executing SHM has its velocity

$10 cm/s$ and

$7 cm/s$ when its displacements from mean position are

$3 cm$ and

$4 cm$ respectively. The length of path is

0%
$10 cm$
0%
$9.5 cm$
0%
$4 cm$
0%
$11.36 cm$

Explanation

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

$v_{1}=10 cm /sec$

$v_{2}=7 cm /sec$

$x_{1}=3 cm$

$x_{2}=4cm$

$v_{1}=\omega\sqrt{A^{2}-x^{2}_{1}}$

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

$\therefore \dfrac{10}{7}=\dfrac{\omega \sqrt{A^{2}-(3)^{2}}}{\omega \sqrt{A^{2}-(4)^{2}}}$

$\dfrac{100}{49}=\dfrac{A^{2}-9}{A^{2}-16}$

$51A^{2}=1600-(9\times 49)$

$A=4.76 cm$
Length of the path is

$2A = 9.52$ cm

A particle executes SHM in a straight line. The maximum speed of the particle during its motion is

$V_{max}$ . Then the average speed of the particle during the SHM is :

0%
$\dfrac{V_{m}}{\pi }$
0%
$\dfrac{V_{m}}{2\pi }$
0%
$\dfrac{2V_{m}}{\pi }$
0%
$\dfrac{3V_{m}}{\pi }$

Explanation

$Vm=A\omega$

$V avg=\dfrac{distance \ \ traveled }{time \ \ taken}$

$=\dfrac{A}{T}$

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

$Vavg=\dfrac{4 A\times \omega}{2\pi} = \dfrac{2 A \times \omega}{\pi}$

$Vavg =\dfrac{2 V_m}{\pi}$

The minimum phase difference between two SHM's is:

$y_{1}=\sin \left(\dfrac{\pi}{6}\right) \sin(\omega t)+\sin\left(\dfrac{\pi }{3}\right) \cos(\omega t )$

$y_{2}=\cos\left(\dfrac{\pi}{6}\right) \sin(\omega t)+\sin \left(\dfrac{\pi}{3}\right) \cos(\omega t)$ is:

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{12}$
0%
$0$

Explanation

$y_{1}=\sin(\omega t)\sin\left(\dfrac{\pi }{6}\right)+ \cos$

$\omega t \sin\dfrac{\pi }{3}$

$=\sin$

$\omega t \cos\left(\dfrac{\pi }{3}\right)+\cos$

$\omega t \sin\dfrac{\pi }{3}$

$=\sin\left(\omega t +\dfrac{\pi }{3}\right)$

Similarly for the second equation is:

$y_{2}=\sin\left(\omega t +\dfrac{\pi }{6}\right)$

Thus phase difference between the two waves is

$\dfrac{\pi }{3}-\dfrac{\pi }{6}=\dfrac{\pi }{6}$

A 1 kg mass executes SHM with an amplitude 10 cm, it takes

$2\pi$ seconds to go from one end to the other end. The magnitude of the force acting on it at any end is :

0%
0.1 N
0%
0.2 N
0%
0.5 N
0%
0.05 N

Explanation

$w = \cfrac{2\pi}{T} = 1 \ rad/sec$ ; Amplitude

$A = 0.1 m$
magnitude of force

$= m \times w^{2}.A$

$= 0.1 N$

A particle free to move along the x-axis has potential energy given by

$U(x)=k[1-exp\left ( -x^{2} \right )]$ for

$-\infty \leq x\leq +\infty ,$ where k is a positive constant of appropriate dimensions. Then:

0%
at points away from the origin, the particle is in unstable equilibrium.
0%
for any finite non-zero value of x, there is a force directed away from the origin.
0%
if its total mechanical energy is k/2 it has its minimum kinetic energy at the origin.
0%
for small displacements from x = 0, the motion is SHM.

Explanation

$exp \left ( -x^{2} \right ) = 1 - x^{2} + \cfrac{x^{4}}{2!} + ---$
For small

$x$ :

$exp \left ( -x^{2} \right ) = (1 - x^{2})$
Thus

$U(x) = K [1-(1-x^{2})] = K x^{2}$

$F = - \dfrac{dU}{dx} = - \dfrac{d(kx^{2})}{dx} = -2 K x$
Thus, the motion is an SHM.

A particle of mass

$m$ is executing oscillations about the origin on the

$x$ -axis. It's potential energy is

$U(x)=k\left | x \right |^{3}$ , where

$k$ is a positive constant. If the amplitude of oscillation is

$a$ , then its time period

$T$ is:

0%
proportional to $\dfrac{1}{\sqrt{a}}$
0%
independent of $a$
0%
proportional to $\sqrt{a}$
0%
proportional to $a^{3/2}$

Explanation

$U=k |x|^3 \Rightarrow F=-\dfrac{dU}{dx}=-3k|x|^2 ...(i)$
Also, for SHM

$x=a \sin \omega t$ and

$\dfrac{d^2x}{dt^2}+\omega ^2x=0$

$\Rightarrow acceleration\ a=\dfrac{d^2x}{dt^2}=-\omega ^2x \Rightarrow F=ma$

$=m \dfrac{d^2x}{dt^2}= - m \omega ^2x ...(ii)$
From equation

$(i)$ &

$(ii)$ we get

$\omega =\sqrt{\dfrac{3kx}{m}}$

$\Rightarrow T=\dfrac{2 \pi}{\omega}= 2 \pi \sqrt{\dfrac{m}{3kx}}= 2 \pi \sqrt{\dfrac{m}{3k(a \sin \omega t)}} \Rightarrow T \propto \dfrac{1}{\sqrt a}$

A body of mass 1/4 kg is in S.H.M and its displacement is given by the relation

$y= 0.05 sin(20t+\dfrac{\pi }{2})$ m. If

$t$ is in seconds, the maximum force acting on the particle is:

0%
$5$ N
0%
$2.5$ N
0%
$10$ N
0%
$0.25$ N

Explanation

We know that

$F= m\omega^{2}A$

$\omega = 20 rad / sec$

$A = 0.05 m$
Thus

$F= \dfrac{1}{4}\times 20\times 20\times \dfrac{1}{20}$

$=5 N$

A body is executing SHM. If the force acting on the body is 6N when the displacement is 2 cm, then the force acting on the body when the displacement is 3 cm in newton is:

0%
$6$ N
0%
$9$ N
0%
$4$ N
0%
$\sqrt{6}$ N

Explanation

$F = m w^{2}x$

$\dfrac{F_{1}}{F_{2}} = \dfrac{x_{1}}{x_{2}}$

$F_{2} = \dfrac{3}{2}\times 6 \ N = 9 \ N$

Two particles are in S.H.M. along parallel straight lines with same amplitude and time period. If they cross each other in opposite directions at the midpoint of mean and extreme positions. Phase difference between them is:

0%
$30^{\circ}$
0%
$120^{\circ}$
0%
$150^{\circ}$
0%
$180^{\circ}$

Explanation

Since

$x=A\sin { \left( \omega t+\phi \right) }$

$\\ \dfrac { A }{ 2 } =A\sin { \left( \omega t+\phi \right) } \\ \Rightarrow \sin { \left( \omega t+\phi \right) } =\frac { 1 }{ 2 } \\ \Rightarrow \omega t+\phi =30^{ \circ }or\quad 150^{ \circ }$
So the one particle has phase of

$30^{ \circ }$ and another has

$150^{ \circ }$
So the phase difference is

$150^{ \circ }-30^{ \circ }=120^{ \circ }$

A disc of mass

$M$ is attached to a horizontal massless spring of force constant

$K$ so that it can roll without slipping along a horizontal surface. If the disc is pushed a little towards right and then released, its center of mass executes SHM with a period of

0%
$2\pi$ $\sqrt{\dfrac{M}{K}}$
0%
$2\pi$ $\sqrt{\dfrac{3M}{K}}$
0%
$2\pi$ $\sqrt{\dfrac{M}{2K}}$
0%
$2\pi$ $\sqrt{\dfrac{3M}{2K}}$

Explanation

Spring exerts a force equal to

$Kx$ on the center of mass of the disc.

Friction acts on opposite direction to cause rolling without slipping.

$Kx-f=Ma_{cm}$

$fR=\dfrac{1}{2}MR^2\alpha$

Solving gives

$f=\dfrac{Kx}{3}$

Hence restoring force acting on sphere=

$kx-f=\dfrac{2Kx}{3}$

Hence

$\omega=\sqrt{\dfrac{2K}{3M}}$

Hence time period of oscillation=

$\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{3M}{2K}}$

A simple harmonic oscillator is of mass

$0.100$ kg. It is oscillating with a frequency of

$\dfrac{5}{\pi }$ Hz. If its amplitude of vibration is

$5$ cm, the force acting on the particle at its extreme position is

0%
2 N
0%
1.5 N
0%
1 N
0%
0.5 N

Explanation

$F = m {\omega}^{2} A$

$\omega = 2\pi f = 10 rad / sec$

$F = 0.1 \times 100\times \dfrac{5}{100}$

$= 0.5 N$

If the displacement (x) and velocity (v) of a particle executing S.H.M. are related through the expression

$4v^2= 25 - x^2$ , then its time period is :

0%
$\pi$
0%
$2 \pi$
0%
$4 \pi$
0%
$6 \pi$

Explanation

Velocity in SHM is given as

$V^{2}=\omega^{2}(A^{2}-x^{2})$
Thus

$V^{2}=\dfrac{1}{4}(5^{2}-x^{2})$

$\therefore \omega =\dfrac{1}{2}$
Thus we get the time period as

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

$T=4\pi sec$

Two particles executing SHM of the same amplitude and frequency on parallel lines side by side. They cross one another when moving in opposite directions each time their displacement is

$\dfrac{\sqrt{3}}{2}$ times their amplitude. The phase difference between them is:

0%
$30^{\circ}$
0%
$45^{\circ}$
0%
$60^{\circ}$
0%
$90^{\circ}$

Explanation

since

$x=A\sin { \left( \omega t+\phi \right) }$

$\\ \dfrac { \sqrt{3}A }{ 2 } =A\sin { \left( \omega t+\phi \right) } \\ \Rightarrow \sin { \left( \omega t+\phi \right) } =\dfrac{\sqrt{3}}{2} \\ \Rightarrow \omega t+\phi =60^{ \circ }or\quad 120^{ \circ }$
so the one particle has phase of

$60^{ \circ }$ and another has

$120^{ \circ }$
So the phase difference is

$120^{ \circ }-60^{ \circ }=60^{ \circ }$

A particle is in S.H.M of amplitude

$2$ cm. At extreme position the force is

$4$ N. At the point mid-way between mean and extreme position, the force is :

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$1$ N
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$2$ N
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$3$ N
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$4$ N

Explanation

Amplitude = 2 cm
Force = 4N

$F = mw^{2}A=4$

$F_{1} = mw^{2}x$

Since

$F$ is directly proportional to

$x$ so , at midpoint the force when the amplitude is

$2 \ cm$ will be

$2N$

The average speed of S.H.M oscillator averaged over

$\dfrac{3T}{8}$ is

0%
$\dfrac{4\sqrt{2}}{3} \dfrac{A\omega }{\pi }$
0%
$\dfrac{2A(4-\sqrt{2})}{3\pi }$
0%
$\dfrac{3A\omega 4- \sqrt{2}\omega }{4\pi }$
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$\dfrac{5\sqrt{2\pi }}{2\omega }$

The net external force acting on the disc when its centre of mass is at displacement

$x$ with respect to its equilibrium position is:

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$-kx$
0%
$-2kx$
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$-\dfrac{2kx}{3}$
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$\dfrac{4kx}{3}$

Explanation

$\textbf{Given:- }$ The net external force acting on the disc

$\textbf{Solution:-}$ Net Force acting on a body is given by the formula:

$\vec{F_{net}} = m.\vec{a}$

Where,

$\vec{F_{net}}$ is the net force acting on the body

m is the mass of the body

$\vec{a}$ is the acceleration of the body

We will insert

$\left ( -2kx + F \right )$ in the place of

$\vec{F_{net}}$

$-2kx + F = Ma_{c}................(1)$

Where,

k is the spring constant

x is the distance by which the spring has been stretched

F is the friction force on the disk

M is the mass of disk

$a_{c}$ is the acceleration of centre of mass of the disk

Net Torque acting on a body is given by the formula,

$\vec{\tau _{net}} = I\vec{\alpha }...............(2)$

Where,

$\vec{\tau _{net}}$ is the net Torque acting on the body

I is the moment of inertia of the body

$\vec{\alpha }$ is the angular acceleration of the body

Additionally, Torque is also calculated as follows,

$\vec{\tau _{net}} = \vec{F}\times R................(3)$

Where,

R is the distance of the Force from the center of mass of the body

Now combining equations 2 and 3,

$\vec{F}\times R = T\vec{\alpha}$

$\vec{F} = \dfrac{T\vec{\alpha}}{R}.................(4)$

In Pure Rolling condition,

$a_{c} = \vec{\alpha_{c}}R$

Moment of inertia of a disk is given by the formula,

$= \dfrac{\dfrac{1}{2}MR^{2}}{R}\times \dfrac{a_{c}}{R}$

$\Rightarrow F = \dfrac{1}{2}Ma_{c}$

Inserting the value of F in equation 1,

$\Rightarrow -2kx + \dfrac{1}{2}Ma_{c} = -Ma_{c}$

$\Rightarrow \dfrac{1}{2}Ma_{c} + Ma_{c} = 2kx$

$\Rightarrow \dfrac{3}{2}Ma_{c} = 2kx$

$\Rightarrow Ma_{c} = \dfrac{4}{3}kx$

$\Rightarrow MA_{c} = \dfrac{4}{3}kx$

$\textbf{Hence the correct option is D}$

A particle of mass 1kg is moving in a S.H.M with an amplitude of 0.02m and a frequency of 60Hz. The maximum force acting on the particle is :

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$2.88 \times 10^{3}$ N
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$1.44 \times 10^{3}$ N
0%
$5.67 \times 10^{3}$ N
0%
$0.75 \times 10^{3}$ N

Explanation

$f = 60\ Hz$

$w = 2\pi f = 120\pi \ rad/sec$

$F = mw^{2}A$

$= 1 \ kg \times 14400 \times \pi ^{2} \times 0.02$

$= 2.88 \times 10^{3}N$

A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is 1 cm, the numerical value of the magnitude of acceleration is equal to the numerical value of the magnitude of the velocity. The frequency of SHM is:

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$2\pi \sqrt{3}$
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$\dfrac{2\pi }{\sqrt{3}}$
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$\dfrac{\sqrt{3}}{2\pi }$
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$\dfrac{1}{2\pi \sqrt{3}}$

Explanation

Let y is the displacement $y=a\sin \omega t$

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

It is given that,

Displacement y = 1

Amplitude a = 2

So $1=3\sin \dfrac{2\pi t}{T}$

Since, acceleration in SHM is $a={{\omega }^{2}}y$

And velocity in SHM is $v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}....(1)$

$\because v=\omega$

Put all the values in equation (1)

$\dfrac{4{{\pi }^{2}}}{{{T}^{2}}}=\dfrac{2\pi }{T}\left( \sqrt{{{a}^{2}}-1} \right)$

$\dfrac{2\pi }{T}=\left( \sqrt{3} \right)$

$T=\dfrac{2\pi }{\sqrt{3}}$

The number of independent constituent simple harmonic motions yielding a resultant displacement equation of the periodic motion as

$y=8 sin^{2}(\frac{t}{2})sin (10t)$ is:

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8
0%
6
0%
4
0%
3

Explanation

The resultant displacement equation of periodic motion is given by

$y=8sin^2(\dfrac{t}{2}). sin(10t)$

$y=4.2sin^2(\dfrac{t}{2}). sin (10t)$

$y=4[1-cost].sin(10t)$ (as,

$2sin^2(\dfrac{t}{2})=1-cost$ )

$y=4sin(10t)-4sin(10t).cost$

$y=4sin(10t)-2sin(11t)-sin(9t)$ (as,

$2sinAtsinBt=sin(A+B)t+sin(A-B)t$ )

There are three independent constituent simple harmonic motion.

The correct option is D.

A particle performs SHM with a period T and amplitude a. The mean velocity of the particle over thetime interval during which it travels a distance a/2 from the extreme position is

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$\dfrac{a}{T}$
0%
$\dfrac{2a}{T}$
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$\dfrac{3a}{T}$
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$\dfrac{a}{2T}$

A particle is subjected to two mutually perpendicular simple harmonic motions such that its x andy coordinates are given by

$x=2 \sin \omega t; y=2 \sin \left ( \omega t+\frac{\pi }{4} \right )$ The path of the particle will be :

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

A spring is suspended under gravity with a block attached to it. The block oscillates in the vertical plane according to

$x = a sin wt$ with time period

$8$ seconds. The ratio of velocities of the particle at

$t = 1$ sec to

$t = 3$ sec is

0%
2 : 1
0%
1 : 1
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3 : 4
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2 : 3

Explanation

$x = a sin \omega t$

$V = \dfrac {dx}{xt} = a\omega cos \omega t$

$T = \dfrac {2\pi}{\omega} ; \omega = \dfrac {2\pi}{T} = \dfrac {\pi}{4} \dfrac {rad}{sec}$

$\dfrac {V_1}{V_2} = \dfrac {cos \omega t_1}{cos \omega t_2} = \dfrac {cos \dfrac {\pi}{4}}{cos \dfrac {3\pi}{4}} = \dfrac {1}{1}$

The mean value of velocity vector projection average over

$3/8$ th of a period after the start is

0%
$\dfrac{2\sqrt{2}A\omega }{\pi }$
0%
$\dfrac{2 A \omega }{3\sqrt{2\pi }}$
0%
$\dfrac{A\omega }{3\pi }$
0%
$\dfrac{2\sqrt{2}A\omega }{3\pi }$

Explanation

$x = A sin\ \omega t$

$V = \dfrac {dx}{dt} = A\omega\ cos \omega t$

$\dfrac {^\dfrac {3T}{8}_{0} \int V dt}{^\dfrac {3T}{8}_{0} \int dt} = \dfrac {\dfrac {A\omega}{\omega}\times [sin \omega t]^{\dfrac {3T}{8}}_{0}}{[t]^{\dfrac {3T}{8}}_{0}}$

$\dfrac {A \times [Sin(\dfrac{2\pi}{T}\times\dfrac {3T}{84})-Sin 0]}{\dfrac{3T}{8}}$

$= \dfrac {A \times \dfrac {1}{\sqrt 2}}{\dfrac {3T}{8}}$

$= \dfrac {2\sqrt 2 A\omega}{3\pi}$

A particle executes simple harmonic oscillation
with an amplitude a. The period of oscillation is
T. The minimum time taken by the particle to
travel half of the amplitude from the
equilibrium position is:

0%
T/2
0%
T/4
0%
T/8
0%
T/12

Explanation

$\because X=asin\omega t$

$\therefore \frac{a}{2}=aisn\omega t \Rightarrow \omega t=\frac{\pi }{6}$

$\Rightarrow \left ( \frac{2\pi }{T} \right )t=\frac{\pi }{6}\Rightarrow t=\frac{T}{12}$

The displacement of a particle along the x-axis is given by x = a

$sin^{2}$ t. The motion of the particle corresponds to

0%
simple harmonic motion of frequency $\omega /\pi$
0%
simple harmonic motion of frequency 3 $\omega /2\pi$
0%
non simple harmonic motion
0%
simple harmonic motion of frequency $\omega /2\pi$

Explanation

$x=asin^{2}t$

$x=\dfrac{a}{2}(1-cos2t)$

from the above equation

$\omega=2$

frequency

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

frequency=

$\dfrac{\omega}{\pi}$

A particle moving in a straight line has, magnitude of velocity

$v$ given by

$v ^{2} = 12 - 3x ^{2}$ , where

$v$ is in m/s and

$x$ is in m. The amplitude of the oscillation of the particle is (in m)

0%
1 m
0%
2 m
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4 m
0%
12 m

Explanation

expression of velocity in S.H.M

$v^2 = 12 - 3 x^{2}$

$v^2 =3( 4 - x^{2})$

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

$A = 2 m$

A particle performing

$SHM$ is found at its equilibrium at

$t=1\ sec$ and it is found to have a speed of

$0.25\ m/s$ at

$t=2\:s$ . If the period of oscillation is

$6\ sec$ , calculate amplitude of oscillation.

0%
$\displaystyle\frac{3}{2\pi}\:m$
0%
$\displaystyle\frac{3}{4\pi}\:m$
0%
$\displaystyle\frac{6}{\pi}\:m$
0%
$\displaystyle\frac{3}{8\pi}\:m$

Explanation

Since a SHM can be represented by

$x=A\sin { \left( \omega t+\phi \right) }$

$\Rightarrow x=A\sin { \left( \left( \dfrac { 2\pi }{ T } \right) t+\phi \right) }$ ;

$T=6s$

so the equation becomes

$x=A\sin { \left( \dfrac { \pi t }{ 3 } +\phi \right) }$

We are given that at

$t=1s, x=0$ ; so we have

$0=A\sin { \left( \dfrac { \pi }{ 3 } +\phi \right) } \\ \Rightarrow \left( \dfrac { \pi }{ 3 } +\phi \right) =0\\ \Rightarrow \phi =-\dfrac { \pi }{ 3 } \\ \Rightarrow x=A\sin { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi }{ 3 } \right) } \\ \Rightarrow v=\dfrac { dx }{ dt } =\dfrac { d }{ dt } \left( A\sin { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi }{ 3 } \right) } \right) =\dfrac { \pi A }{ 3 } \cos { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi }{ 3 } \right) }$
Now we are given that at

$t=2s, \left| v \right| =0.25m/s$ , so we have

$0.25=\left| \dfrac { \pi A }{ 3 } \cos { \left( \dfrac { 2\pi }{ 3 } -\dfrac { \pi }{ 3 } \right) } \right| \\ \Rightarrow 0.25=\left| \dfrac { \pi A }{ 3 } \cos { \left( \dfrac { \pi }{ 3 } \right) } \right| \\ \Rightarrow 0.25=\dfrac { \pi A }{ 6 } \\ \Rightarrow A=\dfrac { 3 }{ 2\pi }$

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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