MCQ Questions for Class 11 Engineering Physics Oscillations Quiz 14

Two particles are executing simple harmonic motion of the same amplitude

$A$ and frequency

$\omega$ along the x-axis. Their mean position is separated by distance

${X}_{0}({X}_{0}> A)$ . If the maximum separation between them is

$({X}_{0}+A)$ , the phase difference between their motion is

0%
$\pi /2$
0%
$\pi /3$
0%
$\pi /4$
0%
$\pi /6$

Explanation

$\begin{array}{l} { x_{ 1 } }=A\sin \left( { \omega t+{ \phi _{ 1 } } } \right) \\ { x_{ 2 } }=A\sin \left( { \omega t+{ \phi _{ 2 } } } \right) \\ { x_{ 1 } }-{ x_{ 2 } }=A\left[ { 2\sin \left[ { \omega t+\dfrac { { { \phi _{ 1 } }+{ \phi _{ 2 } } } }{ 2 } } \right] \sin \left[ { \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } } \right] } \right] \\ A=2A\sin \left( { \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } } \right) \\ \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } =\dfrac { \pi }{ 6 } \\ { \phi _{ 1 } }=\dfrac { \pi }{ 3 } \end{array}$

Hence,

option

$(B)$ is correct answer.

Three measurements of the time for

$20$ oscillations of a pendulum give

$t_1 = 39.6_s , t_2 = 39.9_s and t_3 = 39.5_s$ . The precision in the measurement, is

0%
$0.005 sec$
0%
$0.01 sec$
0%
$1 sec$
0%
$2 sec$

A particle of mass 0.1 kg executes SHM under a force F= (-10x)N. Speed of particle at mean position is 6 m/s. Then amplitude of oscillations is

0%
0.6 m
0%
0.2 m
0%
0.4 m
0%
0.1 m

In the arrangement shown, the solid cylinder of mass m is slightly rolled to the left and released. It starts oscillating on the horizontal surface without slipping. Then time period of oscillation is

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

Explanation

Torque=

$f(R)$

$R\rightarrow$ Radius of cylinder

$I\propto = fR$ moment of inertia for solid cylinder

As no slipping

$I=\dfrac{1}{2}mR^2$

$a=R\propto$

$\dfrac{1}{2} mR^2\left(\dfrac{a}{R}\right)=fR\longrightarrow \therefore f=\dfrac{1}{2}ma$

$ma=-kx-f=-kx-\dfrac{1}{2}ma$

$-kx=\dfrac{3}{2}ma\Rightarrow \dfrac{d^2x}{dt^2}+\dfrac{2}{3}\dfrac{k}{m}x=0$

It is of form

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

$\therefore w=\sqrt{\dfrac{2}{3}\dfrac{k}{m}}$

$T=\dfrac{2\pi}{w}=2\pi \sqrt{\dfrac{3}{2}\dfrac{m}{R}}$

A bob is suspended with the help of a light and rigid rod of length l. A spring of spring consatnt k is also attached. what should be the value of I so the period for small oscillation is

$\pi \sqrt \frac { I } { g }$

0%
$\frac { mg } { k }$
0%
$\frac { 2mg } { k }$
0%
$\frac { 3mg } { k }$
0%
$\frac { 4mg } { k }$

$y= \sin^2\omega t$ represents the displacement of a particle performing SHM, the particle oscillates between

0%
$-1$ and $-\frac { 1} {2}$
0%
$-\frac {1} {2}$ and $-\frac {1} {2}$
0%
$-1$ and $0$
0%
$0$ and $+1$

A particle moves on the

$X-$ axis according to the equation

$x=x_0\sin^{2}\omega t$ . The motion is simple harmonic

0%
With amplitude $x_0$
0%
With amplitude $2\ x_0$
0%
With time period $\dfrac {2\pi}{\omega}$
0%
With time period $\dfrac {\pi}{\omega}$ .

Two particles are in

$SHM$ with same amplitude

$A$ and same angular frequency

$\omega$ . At time

$t=0$ , one is at

$x=+\dfrac{A}{2}$ and other is at

$x=-\dfrac{A}{2}$ . Both are moving in same direction.

0%
Phase difference between the two particle is $\dfrac{\pi}{3}$
0%
Phase difference between the two particle is $\dfrac{2\pi}{3}$
0%
They will collide after time $t=\dfrac{\pi}{2\omega}$
0%
They will collide after time $t=\dfrac{3\pi}{\omega}$

Maximum tension to the minimum tension in the string during oscillations is?(Consider amplitude of oscillation =

$\theta$ )

0%
$(1+\theta^2):(1-\theta^2)$
0%
$(1+\theta^2):(1-\dfrac{\theta^2}{2})$
0%
$(1+\theta^2):1$
0%
$(1+\dfrac{\theta^2}{2}):(1-\theta^2)$

Explanation

At extrem position velocity of ball will be zero.

$T_{min}=Mg\cos\theta$

For velocity at mean position

Form conservation of energy:

Gain in K.E = Loss in PE

$\dfrac{1}{2}mv^2=mg(l-l\cos\theta)$

$\dfrac{mv^2}l=2mg(1-\cos\theta)$

$T=mg+\dfrac{mv^2}{l}=mg+2gm(1-\cos\theta)$

$T_{max}=mg[3-2\cos\theta]$

$\dfrac{T_{max}}{T_{min}}=\dfrac{3-2\cos\theta}{\cos\theta}=\dfrac{3-2\left[1-2\sin^2\dfrac{\theta}{2}\right]}{1-2\sin^2\dfrac{\theta}{2}}$

For small angular displacement

$\dfrac{T_{max}}{T_{min}}=\dfrac{3-2\left[1-\dfrac{\theta^2}{2}\right]}{1-\dfrac{2\theta^2}{4}}=\dfrac{1+\theta^2}{1-\dfrac{\theta^2}{2}}$

1492583_1261132_ans_98c4300f809043c2a0dbf3c8e3763852.png

A particle executes SHM along the line AB . IFC divided AB in the ratio 3 : 1, the ration of the time taken to travel AC and CB is :

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

The period of a conical pendulum of string length

$\sqrt 2$ is

$2 \ s$ . The radius of the bob's orbit is about horizontal is

$[g=9.8 \ m/s^2]$

0%
$\cfrac 1{\sqrt 2} \ m$
0%
$1 \ m$
0%
$\sqrt 2 \ m$
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$\sqrt 3 \ m$

A SHM is given by

$y = 2(sin \omega t + cos \omega t)$ . Which of the following statements are true-

0%
The amplitude is 1 m
0%
The amplitude is $2 \sqrt 2m$
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When t = 0, the amplitude is 0 m
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When t = 0, the amplitude is 2 m

A particle executing SHM while moving from one extreme is found at distance

$x_1,\, x_2$ and

$x_3$ from the centre at the end of three successive seconds. The time period of oscillation is: Here

$\theta = {\cos ^{ - 1}}\left( {\frac{{{x_1} + {x_3}}}{{2{x_2}}}} \right)$

0%
$\dfrac {2 \pi}{\theta}$
0%
$\dfrac { \pi}{\theta}$
0%
$\theta$
0%
$\dfrac { \pi}{2\theta}$

A block of mass

$M$ is performing

$SHM$ with amplitude

$A$ on a smooth horizontal surface

$.$ At the extreme position a small block of mass

$m$ falls vertically and sticks to

$M.$ then

$,$ amplitude of oscillation will be

0%
$A$
0%
$A\sqrt {\dfrac{M}{{M + m}}}$
0%
$A\left( {\dfrac{M}{{M + m}}} \right)$
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$A\left( {\dfrac{{M + m}}{m}} \right)$

A particle of mass

$m$ is allowed to oscillate on a smooth parabola:

${ x }^{ 2 }=4ay,a>1$ as shown in the figure. The angular frequency

$\left( \omega \right)$ of small oscillations is

0%
$\omega =\sqrt { \dfrac { g }{ 4a } }$
0%
$\omega =\sqrt { \dfrac { g }{ 2a } }$
0%
$\omega =\sqrt { \dfrac { 2g }{ a } }$
0%
$\omega =\sqrt { \dfrac { g }{ a } }$

Explanation

Ref.Image.

$x^{2} = 4ay$

$2x = 4a \dfrac{dy}{dx}$

$tan\theta = \dfrac{dy}{dx} = \dfrac{2x}{4a}$

fresing = -mg

$sin\theta$

$tan\theta -sin\theta \simeq \theta .tan\theta$ is very small

$tan\theta = \dfrac{x}{20}$

$F=-mg \dfrac{X}{20}$

$F = -k/l$

$K= \dfrac{mg}{2a} K = mw^{2}$

$\dfrac{mg}{2a} = mw^{2}$

$w2 \sqrt{\dfrac{y}{2a}}$

1240380_1303958_ans_6aa995de03894579976c55743cf13d81.png

If two wires of same length

$\ell$ and area of cross section A with young modulus y and 2y connect in series and one end is fixed on roof and other end with mass m. Make simple harmonic motion, then the time period is-

0%
$2\pi \sqrt { \dfrac { m\ell }{ YA } }$
0%
$2\pi \sqrt { \dfrac { m\ell }{ 3YA } }$
0%
$2\pi \sqrt { \dfrac { 3m\ell }{ 2YA } }$
0%
$2\pi \sqrt { \dfrac { m\ell }{ 2YA } }$

A linear harmonic oscillator of force constant $6 \times 10^5 N/m$ and amplitude $4$ cm, has total energy $600 J$ . Select the correct statement.

0%
Maximum potential energy is $600$ J
0%
Maximum kinetic energy is $480$ J
0%
Minimum potential energy is $120$ J
0%
None

A particle executing S.H.M. given by equation

$y= 8 sin 6 \pi t$ is sending out waves in a continuous medium traveling at 200 cm/s. the resultant displacement of the particle 150 cm, from B and one second after commencement of vibration of B is:

0%
4 cm
0%
8 cm
0%
-8 cm
0%
-3 cm

Equation of SHM is x = 10 sin 10

$\pi t$ . Find the distance between the two points where speed is

$50\pi$ m/s. x is in cm and t is in seconds.

0%
10 cm
0%
14 cm
0%
17.32 cm
0%
8.66 cm

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency off 1 rad/s and an amplitude of

${ 10 }^{ -2 }.$ The relative change in the angular frequency of the pendulum is best given by:

0%
${ 10 }^{ -5 }rad/s$
0%
${ 10 }^{ -1 }rad/s$
0%
${ 10 }^{ }rad/s$
0%
${ 10 }^{ -3 }rad/s$

As shown in figure a horizontal platform with a mass

$m$ placed on it is executing SHM along y-axis. If the amplitude of oscillation is

$2.5cm$ , the minimum period of the motion for the mass not to be detached from the platform is
(

$g=10m/{sec}^{2}={\pi}^{2}$ )

0%
$\cfrac{10}{\pi}s$
0%
$\cfrac{\pi}{10}s$
0%
$\cfrac{\pi}{\sqrt{10}}s$
0%
$\cfrac{1}{\sqrt{10}}s$

Two particles are in SHM in a straight line about same equilibrium position. Amplitude

$A$ and time period

$T$ of both the particles are equal. At time

$t = 0$ , one particle is at displacement

$y_{1} = +A$ and the other at

$y_{2} = -A/2$ , and they are approaching towards towards each other. After what time they cross each other?

0%
$T/3$
0%
$T/4$
0%
$5T/6$
0%
$T/6$

A thin hoop of radius r and mass m is suspended from a rough rod as shown. Determine the time period of small oscillaion of the hoop in a direction perpendicular to the plane of the hoop. Assume that friction is sufficiently large to prevent slipping at A.

0%
$2 \pi \sqrt {\cfrac g R}$
0%
$2 \pi \sqrt {\cfrac {3R} {2g}}$
0%
$2 \pi \sqrt {\cfrac R {2g}}$
0%
None

A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of

$40\ cm$ . The block just loses contact with the plank when the latter is at momentary rest. Then

0%
the period of oscillation is $\left( \dfrac { 2\pi }{ 5 } \right)$
0%
the block weight double its weight, when the plank is at one of the position of momentary rest
0%
the block weighs 0.5 times its weight on the plank halfway up
0%
the block weighs 1.5 times its weight on the plank halfway down.

$ABC$ is an equilateral triangle structure up of a light rigid material. Find the frequency of small vertical oscillations of mass

$m$ along

$AG$ . Conisider

${k}_{1}={k}_{2}={k}_{3}={k}_{4}=k$

0%
$\sqrt{\cfrac{5k}{2m}}$
0%
$\sqrt{\cfrac{4k}{m}}$
0%
$\sqrt{\cfrac{3k}{5m}}$
0%
$\sqrt{\cfrac{10k}{9m}}$

The length of simple pendulum is about 100 cm known to have an accuracy of 1 mm. Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a block of 0.1 s resolution. What is the accuracy in the determined value g?

0%
0.2%
0%
0.5%
0%
0.1%
0%
2%

A spring is placed in vertical position by suspending it from a hook at its top. A similar hook on the bottom of the spring is at

$11\ cm$ above a table top. A mass of

$75\ g$ and of negligible size is then suspended from the bottom hook, which is measured to be

$4.5\ cm$ above the table top. The mass is then pulled down a distance of

$4\ cm$ and released. Find the approximate position of the bottom hook after

$s$ ?
Take

$g=10m/{s}^{2}$ and hooks mass to be negligible.

0%
$5cm$ above the table top
0%
$4.5cm$ above the table top
0%
$9cm$ above the table top
0%
$0.5cm$ above the table top

A particle executes simple harmonic motion and is located at

$x = a, b$ and

$c$ at times

$t_{0}, 2t_{0}$ and

$3t_{0}$ respectively. The frequency of the oscillation is

0%
$\dfrac {1}{2\pi t_{0}}\cos^{-1} \left (\dfrac {a + c}{2b}\right )$
0%
$\dfrac {1}{2\pi t_{0}}\cos^{-1} \left (\dfrac {a + b}{2c}\right )$
0%
$\dfrac {1}{2\pi t_{0}}\cos^{-1} \left (\dfrac {2a + 3c}{2b}\right )$
0%
$\dfrac {1}{2\pi t_{0}}\cos^{-1} \left (\dfrac {a + 2b}{3c}\right )$

$1.8$ g mass suspended by a spring with a spring constant of

$3$ N/m is forced to oscillate in viscous medium (b=

$2$ g/s) by a driving force of F=

${10^{ - 3}}$ sin

$40$ t (in SI units). the amplitude of the driven oscillations will be

0%
$3.4 \times {10^{ - 6}}{m^2}$
0%
$3.6 \times {10^{ 5}}{m^2}$
0%
$3.4 \times {10^{ 3}}{m^2}$
0%
$2.4 \times {10^{ 2}}{m^2}$

If a body of mass 0.98 kg is made to oscillate on a spring of force constant 4.84 N/m, the angular frequency of the body is

0%
1.22 rad/s
0%
2.22 rad/s
0%
3.22 rad/s
0%
4.22 rad/s

The time period of particle performing linear SHM is 12 s what is time taken by it to cover distance equal to half its amplitude starting its motion from the mean position?

0%
1sec
0%
2sec
0%
3sec
0%
4sec

If the maximum speed of a particle in SHM is

$5\ m/s$ . The average speed of the particle is SHM is equal to

0%
$\dfrac {5}{\pi}m/s$
0%
$\dfrac {10}{\pi}m/s$
0%
$\dfrac {5}{2}m/s$
0%
Zero

Frequency

$f$ of a simple pendulum depends on its length

$\ell$ and acceleration

$g$ due to gravity according to the following equation

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}}$ . Graph between which of the following quantities is a straight line ?

0%
$f$ on the ordinate and $\ell$ on the abcissa
0%
$f$ on the ordinate and $\sqrt{\ell}$ on the abcissa
0%
$f^{2}$ on the ordinate and $\ell$ on the abcissa
0%
$f^{2}$ on the ordinate and $1/\ell$ on the abcissa

A particle is executing SHM with time period T.Starting from mean position, time taken by it to complete

$cfrac 5 8$ oscillations, is

0%
$\cfrac T {12}$
0%
$\cfrac T {6}$
0%
$\cfrac {5T} {12}$
0%
$\cfrac {7T} {12}$

If velocity of a particle in SHM at x=4 m and x=5 m are 15 m/s and 13 m/s then its time period will be:

0%
$\pi$ /4
0%
$\pi$ /2
0%
$\pi$
0%
4 $\pi$ /5

The motion represented by equation

$x=2\sin \omega t+3\sin^{2}\omega t$ is

0%
Periodic
0%
Oscillatory
0%
$SHM$
0%
Both (1) & (2)

In a simple oscillating pendulum, the work done by the string in one oscillation will be

0%
Equal to the total energy of the pendulum
0%
Equal to the $K.E$ of the pendulum
0%
Equal to the $P.E$ of the pendulum
0%
Zero

An iron ball of mass

$M$ is hanged from the ceiling by a spring with a spring constant

$k$ . It executes a

$SHM$ with a period

$P$ . If the mass of the ball is increased by four times, the new period will be

0%
$4\ P$
0%
$\dfrac{P}{4}$
0%
$2\ P$
0%
$P$

Initially mass m is held such that spring is in relaxed condition . if mass m is suddenly released maximum elongation in spring will be

0%
$\dfrac {mg}{k}$
0%
$\dfrac {2mg}{k}$
0%
$\dfrac {mg}{2k}$
0%
$\dfrac {mg}{4k}$

Aa particle executes simple harmonic motion with a speed of

$T\ s$ and magnitude

$A\ m$ . The shortest time it takes to reach a point

$\dfrac{A}{\sqrt{2}}\ m$ from its mean position in seconds is:

0%
$T$
0%
$T/4$
0%
$T/8$
0%
$T/16$

A pendulum bob of weight IN is beld at an angle

$\theta$ from the vertical by a horizontal force of 2

$N$ ashown. The tension in the string supporting the pendulum bob (in newton) is

0%
$\frac { 1 } { \cos \theta }$
0%
$\frac { 2 } { 2 \sin \theta }$
0%
$\sqrt { 15 }$
0%
1

A pendulum clock keeps correct time at

$20^{o}C$ . The correction to be made during summer per day, when the average temperature is

$40^{o}C, (\alpha =10^{-5}/^{o}C)$ will be:

0%
$5.64\ sec$
0%
$6.64\ sec$
0%
$7.64\ sec$
0%
$8.64\ sec$

Two pendulums have time periods

$T$ and 5

$T / 4 .$ They start SHM at the same time from the mean position. After how many oscillations of the smaller pendulum they will be again in the same phase:

0%
$5$
0%
$4$
0%
$11$
0%
$9$

A particle executes simple harmonic motion of time period T. At t=0, it is at rest and moves towards positive amplitude then :-

0%
Particle initial phase will be ${ 0 }^{ \circ }$
0%
Particle velocity will be zero after T/2 time interval.
0%
Particle's velocity becomes half of the maximum velocity first time after T/12.
0%
Particle initial phase will be $\left( -\dfrac { \pi }{ 2 } \right)$

A block of mass m moving with speed v compresses a spring through distance X before its speed is halved.What is the value of spring constant?

0%
$\dfrac{{3m{v^2}}}{{4{x^2}}}$
0%
$\dfrac{{m{v^2}}}{{4{x^2}}}$
0%
$\dfrac{{m{v^2}}}{{2{x^2}}}$
0%
$\dfrac{{2m{v^2}}}{{{x^2}}}$

Two particles

$P$ and

$Q$ describe

$S.H.M.$ of same amplitude

$a$ , same frequency

$f$ (this is not angular frequency

$\omega=\dfrac{2\pi}{T}$ but it is

$f=\dfrac{1}{T}$ ) along the same straight line. The maximum distance between two particles is

$a\sqrt{2}$ . The phase difference between the particles is:

0%
$Zero$
0%
$\pi/2$
0%
$\pi/6$
0%
$\pi/3$

A pendulum of mass

$1{ kg }$ and length

$l = 1 { m }$ is released from rest at angle

$\theta = 60 ^ { \circ }.$ The power delivered by all the forces acting on the bob at angle

$\theta = 60 ^ { \circ }$ will be

$\left( { g } = 10 { m } / { s } ^ { 2 } \right).$

0%
$13.4 W$
0%
$20.4 W$
0%
$26.6 W$
0%
zero

Which of the following expressions does not represent Simple Harmonic Motion?

0%
$A \cos \omega t$
0%
$A \sin \omega t$
0%
$A \sin \omega t+B \cos \omega t$
0%
$Ae^{\sin \omega\ t}$

A bullet of mass 10 g moving horizontally with a velocity of

$4 400 ms^{-1}$ $ strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5 m. as a result, the centre of gravity of the block found to rise a vertical distance of 10 cm . the speed of the bullet after it emerges out horizontally from the block will be

0%
$100 ms^{-1}$
0%
$80 ms^{-1}$
0%
$120 ms^{-1}$
0%
$160 ms^{-1}$

Two blocks of masses

$m_{1}=m$ and

$m_{2}3\ m$ are connected by a spring of force constant

$k$ and placed on a horizontal frictionless surface as shown in the fig. The spring is stretched by an amount

$x$ and released. The system executes simple harmonic motion. The relative velocity of the blocks when the spring is at its natural length is

0%
$x \sqrt{\dfrac{3\ k}{2\ m}}$
0%
$2x \sqrt{\dfrac{k}{m}}$
0%
$\dfrac{x}{2} \sqrt{\dfrac{k}{2\ m}}$
0%
$2x \sqrt{\dfrac{k}{3\ m}}$

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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