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

The equation of motion of a particle is $$x = a \cos (a t)^{2}$$. The motion is 
  • periodic but not oscillatory.
  • Periodic and oscillatory.
  • oscillatory but not periodic
  • neither periodic nor oscillatory
The displacement of a particle varies with time, according to the equation $$y = a \sin \omega  t + b \cos \omega t$$
  • The motion is oscillatory but not SHM.
  • The motion is SHM. with amplitude (a + b)
  • The motion is SHM. with amplitude $$(a^{2} + b^{2})$$
  • The motion is SHM. with amplitude $$\sqrt{a^{2} + b^{2}}$$
If a simple pendulum is taken to place where $$g$$ decreases by $$2\%$$ then the time period
  • Decreases by $$1\%$$
  • Increases by $$2\%$$
  • Increase by $$2\%$$
  • Increases by $$1\%$$
A simple pendulum is vibrating in an evacuated chamber, it will oscillate with
  • Increasing amplitude
  • Constant amplitude
  • Decreasing amplitude
  • First $$(c)$$ then $$(a)$$
The velocity of a particle performing simple harmonic motion, when it passes through its mean position is 
  • Infinity
  • Zero
  • Minimum
  • Maximum
The length of simple pendulum is increased by $$1\%$$. Its period will
  • Increase by $$1\%$$
  • Increase by $$0.5\%$$
  • Decrease by $$0.5\%$$
  • Increase by $$2\%$$
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
  • 1 m$$ s^{-1} $$
  • 1.414m$$ s^{-1} $$
  • 0.5m$$ s^{-1} $$
  • 0.75m$$ s^{-1} $$
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 
  • 2 $$ \pi $$cm/s
  • 3 $$ \pi $$ cm/s
  • 4 $$ \pi $$ cm/s
  • 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 :
  • $$ \frac{\pi }{2} +\phi $$
  • $$ - \phi $$
  • $$ \phi $$
  • $$ \phi -\frac{\pi }{2} $$
$$ T _{1}$$, $$ T _{2}$$ 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
  • $$ T_{1}+T_{2} $$
  • $$ \dfrac{T_{1}+T_{2}}{2} $$
  • $$ \dfrac{T_{1}T_{2}}{T_{1}+T_{2}} $$
  • $$ \dfrac{T_{1}T_{2}}{\sqrt{{T_{1}}^{2}+{T_{2}}^{2}}} $$
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:                           
  • $$3$$
  • $$2$$
  • $$1.5$$
  • $$1.0$$
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.
  • Both A and R are true and R is the correct explanation of A
  • Both A and R are true and R is not the correct explanation of A
  • A is true and R is false
  • A is false and R is true
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
  • 1
  • 2
  • 5
  • 6
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:
  • 2$$ \pi \sqrt{B} $$
  • 2$$ \pi / B $$
  • 2$$ \pi/\sqrt{B} $$
  • 2 $$ \pi B $$
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
  • 0.1 m/s
  • 0.2 m/s
  • 0.3 m/s
  • 0.5 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 
  • $$0.1$$ s
  • $$0.2$$ s
  • $$0.4$$ s
  • $$0.6$$ s
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 :
  • $$0.1\ s$$
  • $$0.2\ s$$
  • $$0.3\ s$$
  • $$0.4\ s$$
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:
  • $$\pi $$s
  • 2$$\pi $$s
  • 3$$\pi $$s
  • 4$$\pi $$s
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:
  • $$12/25$$
  • $$24/25$$
  • $$25/24$$
  • $$25/12$$
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
  • $$ \pi/2 $$ s
  • $$ \pi $$ s
  • $$2 \pi $$ s
  • $$ \pi/4 $$ s
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:
  • $$2.25$$ cm/s
  • $$2.5$$ cm/s
  • $$3.0$$ cm/s
  • $$3.5$$ cm/s
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
  • 2:1
  • 1:2
  • 4:3
  • 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 :
  • $$25\ m$$ and $$5\sqrt{3}\ m/s $$ 
  • $$15\ m$$ and $$5\sqrt{3}\ m/s $$
  • $$15\sqrt{3}\ m $$ and $$30\  m/s$$
  • $$ 10\sqrt{3}/2\ m $$ and $$30\ m/s$$
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
  • $$10 cm$$
  • $$9.5 cm$$
  • $$4 cm$$
  • $$11.36 cm$$
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 :
  • $$\dfrac{V_{m}}{\pi }$$
  • $$\dfrac{V_{m}}{2\pi }$$
  • $$\dfrac{2V_{m}}{\pi }$$
  • $$\dfrac{3V_{m}}{\pi }$$
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:
  • $$\dfrac{\pi}{3} $$
  • $$\dfrac{\pi}{6} $$
  • $$\dfrac{\pi}{12} $$
  • $$0$$
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 :
  • 0.1 N
  • 0.2 N
  • 0.5 N
  • 0.05 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:
  • at points away from the origin, the particle is in unstable equilibrium.
  • for any finite non-zero value of x, there is a force directed away from the origin.
  • if its total mechanical energy is k/2 it has its minimum kinetic energy at the origin.
  • for small displacements from x = 0, the motion is 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:
  • proportional to $$\dfrac{1}{\sqrt{a}} $$
  • independent of $$a$$
  • proportional to $$\sqrt{a}$$
  • proportional to $$a^{3/2}$$
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:
  • $$5$$ N
  • $$2.5$$ N
  • $$10$$ N
  • $$0.25$$ 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:
  • $$ 6 $$ N
  • $$9$$ N
  • $$4$$ N
  • $$\sqrt{6} $$ 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:
  • $$30^{\circ}$$
  • $$ 120^{\circ}$$
  • $$150^{\circ}$$
  • $$180^{\circ}$$
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
  • $$2\pi $$ $$\sqrt{\dfrac{M}{K}}$$
  • $$2\pi $$ $$\sqrt{\dfrac{3M}{K}}$$
  • $$2\pi $$ $$\sqrt{\dfrac{M}{2K}}$$
  • $$2\pi $$ $$\sqrt{\dfrac{3M}{2K}}$$
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
  • 2 N
  • 1.5 N
  • 1 N
  • 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 :
  • $$\pi$$
  • $$ 2 \pi $$
  • $$4 \pi$$
  • $$ 6 \pi$$
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:
  • $$30^{\circ}$$
  • $$45^{\circ}$$
  • $$60^{\circ}$$
  • $$90^{\circ}$$
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 :
  • $$1$$ N
  • $$2$$N
  • $$3$$N
  • $$4$$N
The average speed of S.H.M oscillator averaged over  $$\dfrac{3T}{8}$$ is
  • $$\dfrac{4\sqrt{2}}{3} \dfrac{A\omega }{\pi }$$
  • $$\dfrac{2A(4-\sqrt{2})}{3\pi }$$
  • $$\dfrac{3A\omega 4- \sqrt{2}\omega }{4\pi }$$
  • $$\dfrac{5\sqrt{2\pi }}{2\omega }$$
The net external force acting on the disc when its centre of mass is at displacement $$x$$ with respect to its equilibrium position is:
  • $$-kx$$
  • $$-2kx$$
  • $$-\dfrac{2kx}{3}$$
  • $$\dfrac{4kx}{3}$$
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 :
  • $$2.88 \times 10^{3}$$N
  • $$1.44 \times 10^{3}$$N
  • $$5.67 \times 10^{3}$$N
  • $$0.75 \times 10^{3}$$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:
  • $$2\pi \sqrt{3}$$
  • $$\dfrac{2\pi }{\sqrt{3}}$$
  • $$\dfrac{\sqrt{3}}{2\pi }$$
  • $$\dfrac{1}{2\pi \sqrt{3}}$$
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:
  • 8
  • 6
  • 4
  • 3
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
  • $$\dfrac{a}{T}$$
  • $$\dfrac{2a}{T}$$
  • $$\dfrac{3a}{T}$$
  • $$\dfrac{a}{2T}$$
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 :
  • an ellipse
  • a straight line
  • a parabola
  • 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
  • 2 : 1
  • 1 : 1
  • 3 : 4
  • 2 : 3
The mean value of velocity vector projection average over $$3/8$$th of a period after the start is
  • $$\dfrac{2\sqrt{2}A\omega }{\pi }$$
  • $$\dfrac{2 A \omega }{3\sqrt{2\pi }}$$
  • $$\dfrac{A\omega }{3\pi }$$
  • $$\dfrac{2\sqrt{2}A\omega }{3\pi }$$
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:
  • T/2
  • T/4
  • T/8
  • 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
  • simple harmonic motion of frequency $$\omega /\pi $$
  • simple harmonic motion of frequency 3$$\omega /2\pi $$
  • non simple harmonic motion
  • simple harmonic motion of frequency $$\omega /2\pi $$
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)
  • 1 m
  • 2 m
  • 4 m
  • 12 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.
  • $$\displaystyle\frac{3}{2\pi}\:m$$
  • $$\displaystyle\frac{3}{4\pi}\:m$$
  • $$\displaystyle\frac{6}{\pi}\:m$$
  • $$\displaystyle\frac{3}{8\pi}\:m$$
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