CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 9 - MCQExams.com

In the arrangement as shown, the length of rod is L and mass is M and it is connected to smooth pin at O.The rod is placed on smooth horizontal table.The spring constant of both the spring is K.The rod is displaced by a small angle $$\theta$$ to perform small ocillation, then is given by
1237212_81a510815dad42cbbae76a8ff228cc50.png
  • $$\left( \cfrac { 3K }{ M } \right) ^{ 1/2 }$$
  • $$\left( \cfrac { 6K }{ M } \right) ^{ 1/2 }$$
  • $$\left( \cfrac { 4K }{ M } \right) ^{ 1/2 }$$
  • $$\left( \cfrac { 3K }{ 2M } \right) ^{ 1/2 }$$
A simple harmonic wave in a medium is given by,(y is in centimeters) $$ y=\frac { 10 }{ \pi  } sin (2000{ \pi  }t- \frac { \pi x }{ 17 } )$$
The maximum velcocity of a particle executing tha wave is
  • $$330ms^-1$$
  • $$20ms^-1$$
  • $$200ms^-1$$
  • $$2000ms^-1$$
A mass is suspended from a wire and is pulled along the length of wire, resulting in oscillations of time period $$T_{1}$$. The same mass is next attached to wire of the same material and length but double the cross-sectional area. The time period this time $$T_{2}$$. Then $$T_{1}/T_{2}$$ is equal to
  • $$1 : 2$$
  • $$2 : 1$$
  • $$1 : \sqrt{2}$$
  • $$\sqrt {2} : 1$$
Oscillatory or vibrating motion means to and fro motion
  • Which does not repeat
  • which repeats after unequal interval of time
  • which repeats after equal interval of time
  • which remains unchanged
Two particles $$P$$ and $$Q$$ start from origin and execute simple harmonic motion along $$X-$$ axis with same amplitude but with periods $$3\  s$$ and $$6\  s$$ respectively. The ratio of the velocities of $$P$$ and $$Q$$ when they meet is 
  • $$1 : 2$$
  • $$2 : 1$$
  • $$2 : 3$$
  • $$3 : 2$$
Time period of small oscillation (in a vertical plane normal to the plane of strings) of the bob in the arrangement shown will be

1276744_f935a00539c346179924ab1f2a6c7caa.PNG
  • $$2\pi \sqrt{ \cfrac{l}{g}}$$
  • $$2\pi \sqrt {\cfrac {l}{\sqrt {2g}}}$$
  • $$2\pi \sqrt {\cfrac {\sqrt {2l}}{g}}$$
  • $$2\pi \sqrt {\cfrac {2l}{g}}$$
A particle is executing SHM about $$\gamma =0$$ along y- axis.Its position at an instant is given by $$\gamma =(7m)$$ sin(πt) its average velcocity for a time interval 0 to 0.5 s  is
  • $$14m/s$$
  • $$7m/s$$
  • $$\dfrac { 1}{ 7 } m/s$$
  • $$28m/s$$
The velocity v of a particle of mass m moving along a straight line changes with time 't' as $$\dfrac{{{d^2}v}}{{d{t^2}}} = Kv\;$$ where 'K' is a positive constant. Which of the following statement is correct:
  • The particle does not perform SHM
  • The particle performs SHM with time period $$2\pi \sqrt {\dfrac{m}{k}} $$
  • nothing occurs
  • its in non moving state
The time period of a particle executing $$SHM$$ is $$8\ s$$. At $$t = 0$$ it is at the mean position. The ratio of distance covered by the particle in $${ 1 }^{ st }$$ second to the $${ 2 }^{ nd }$$ second is
  • $$(\dfrac{1}{\sqrt { 2 } -1})$$
  • $$\sqrt { 2 } $$
  • $$(\sqrt { 2 } +1)$$
  • $$\dfrac { 1 }{ \sqrt { 2 } } $$
Initially the spring is undeformed. Now the force $$F$$ is applied to $$B$$ as shown. When the displacement of $$B$$ w.r.t $$A$$ is $$x$$ towards right in some time then the relative acceleration of $$B$$ w.r.t $$A$$ at the moment is:
1272842_03e027574020464f9cff8e5bbb52b3f9.png
  • $$\cfrac{F}{2m}$$
  • $$\cfrac{F-kx}{m}$$
  • $$\cfrac{F-2kx}{m}$$
  • $$none\ of\ these$$
Two particles of same period $$(T)$$ and amplitude undergo $$SHM$$ along the same line with initial phase of $$\pi/6$$. If they start at the same instant and at same point along opposite directions, find the time after which they will meet again for the first time:
  • $$3T/8$$
  • $$T/4$$
  • $$T/2$$
  • $$T$$
The total energy of a particle executing SHM is directly proportional to the square of the following quantity.
  • Acceleration
  • Amplitude
  • Time period
  • Mass
A particle executes $$SHM$$ and its position with time as $$x = A \sin\omega t$$. Its average speed during its motion from mean position to mid-point of mean and extreme position is
  • $$Zero$$
  • $$\cfrac{3A\omega}{\pi}$$
  • $$\cfrac{A\omega}{2\pi}$$
  • $$\cfrac{2A\omega}{\pi}$$

A particle executes SHM according to equation $$x = 10 \cos[2\pi t+\pi / 2]$$, where $$t$$ is in second. The magnitude of the velocity of the particle at $$t=1/6s$$ will be:
  • $$2.47$$ 
  • $$2.86$$ 
  • $$2.05$$ 
  • $$31.4$$

Two charges each $$+5\mu C$$ are at $$(0, \pm 3)$$. A third charge of $$1\mu C$$ and of mass $$2g$$ is at $$(4, 0)$$. What is the minimum velocity to be given to the $$1\mu C$$ charge such that it just reaches the origin (in $$ms^{-1}$$)?
  • $$3$$
  • $$12$$
  • $$\sqrt12$$
  • $$4$$
A particle undergoes $$SHM$$ of period $$T$$. The time taken to complete $$3/8th$$ oscillation starting from the mean position is 
  • $$\dfrac{3}{8}T$$
  • $$\dfrac{5}{8}T$$
  • $$\dfrac{5}{12}T$$
  • $$\dfrac{7}{12}T$$
The mass of particle executing S.H.M is 1 gm.If its periodic time is $$\pi $$ seconds, the value of force constant is:-
  • 4 dynes/cm
  • 4 N/cm
  • 4 N/m
  • 4 dynes/m
A particle is executing SHM about y=0 along y-axis. Its position at an instant is given by y=(7m) sin $$\left( \pi f \right) .$$ its average velocity for a time interval 0 to 0.5 s is
  • 14 m/s
  • 7 m/s
  • $$\dfrac { 1 }{ 7 } m/s$$
  • 28 m/s
A small body of mass $$0.10 kg$$ is executing S.H.M of amplitude $$1.0m$$ and period $$0.20sec$$. The maximum force acting on it is:
  • $$98.596N$$
  • $$985.96N$$
  • $$100.2N$$
  • $$76.23N$$
A particle is executing S.H.M.with amplitude $$5$$ cm along x axis, origin as mean position. If at $$x= +4$$ cm magnitude of velocity is equal to magnitude of acceleration then find time period of oscillation in seconds.
  • $$\frac{4\pi}{3}$$
  • $$6\pi$$
  • $$\frac{8\pi}{3}$$
  • $$\frac{9\pi}{2}$$

A simple harmonic oscillator of angular frequency $$2$$ rad/s is acted upon by an external force $$F = \sin t$$ N. If the oscillator is at rest in its equilibrium position at $$t= 0$$, its position at later times is proportional to:
  • $$ \sin t + \cfrac{1}{2} \cos 2t$$
  • $$ \cos t - \cfrac{1}{2} \sin 2t$$
  • $$ \sin t + \cfrac{1}{2} \sin 2t$$
  • $$ \sin t - \cfrac{1}{2} \sin 2t$$
A student says that he had applied a force $$F= -k\sqrt {x}$$ on a particle and the particle moves in simple harmonic motion. He refuses to tell whether $$k$$ is a constant or not. Assume that he has worked only with positive $$x$$ and no other force acted on the particle.
  • As $$x$$ increases $$k$$ increases
  • As $$x$$ increases $$k$$ decreases
  • As $$x$$ increases $$k$$ remains constant
  • The motion cannot be simple harmonic
If amplitude of particle executing $$SHM$$ is doubled, which of the following quantities are doubled
i) Time period
ii) Maximum velocity
iii) Maximum acceleration
iv) Total energy
  • ii & iii
  • i, ii & iii
  • i, & iii
  • i, ii, iii & iv
A simple harmonic motion has an amplitude A and time period T. Find the time required by it to travel diameter from .
  • $$x = x\,to\,x = \,A/2$$
  • $$x\, = \,0\,\,to\,x\, = \,\frac{A}{{\sqrt 2 }}$$
  • $$x = A\,to\,x = \,A/2$$
  • $$x = - \frac{A}{{\sqrt 2 }}\,to\,x = \,\frac{A}{{\sqrt 2 }}$$
A simple motion is represented by:
$$y = 5(\sin 3 \pi t + \sqrt{3} \cos 3\pi t)cm$$
The amplitude and time period of the motion are:
  • $$5 \,cm, \dfrac{3}{2}s$$
  • $$5 \,cm, \dfrac{2}{3} s$$
  • $$10 \,cm, \dfrac{3}{2}s$$
  • $$10 \,cm, \dfrac{2}{3}s$$
Which of the following quantity is unitless
  • Velocity gradient
  • Pressure gradient
  • Displacement gradient
  • NONE
A block of mass m rests on a platform. The platform is given up and down SHM with an amplitude d- What can be the maximum frequency so that the block never leaves the platform?
  • $$\sqrt { \dfrac { g } { d } }$$
  • $$\dfrac { 1 } { 2 \pi } \sqrt { \dfrac { g } { d } }$$
  • $$\dfrac { 1 } { 2 \pi } \left( \dfrac { g } { d } \right)$$
  • $$2 \pi \sqrt { \dfrac { d } { g } }$$
A particle executes simple harmonic motion with an amplitude of 5 cm. when the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. then, its periodic time in second is:
  • $$\dfrac{7}{3}\pi$$
  • $$\dfrac{3}{8}\pi$$
  • $$\dfrac{4\pi}{3}$$
  • $$\dfrac{8\pi}{3}$$
Two particles undergoing simple harmonic motion of same frequency and same amplitude cross each other at $$x=\dfrac {A}{2}$$. Phase difference between them is
  • $$\dfrac {\pi}{3}$$
  • $$\dfrac {\pi}{2}$$
  • $$\dfrac {2\pi}{3}$$
  • $$\dfrac {\pi}{4}$$
The amplitude $$(A)$$ of damped oscillator becomes half in $$5$$ minutes. The amplitude after next $$10$$ minutes will be:
  • $$A$$
  • $$\cfrac{A}{8}$$
  • $$\cfrac{A}{4}$$
  • $$4A$$
After charging a capacitor $$C$$ to a potential $$V$$ , it is connected across an ideal inductor $$L$$.The capacitor starts discharging simple harmonically at time $$t = 0 .$$The charge on the capacitor at a later time instant is $$q$$ and the periodic time of simple harmonic oscillations is $$T$$. Therefore, 
  • $$q = C V \sin ( \omega t )$$
  • $$q = C V \cos ( \omega t )$$
  • $$T = 2 \pi \sqrt { \frac { 1 } { L C } }$$
  • $$T = 2 \pi \sqrt { L C }$$
A simple pendulum hanging from the ceiling of a stationary lift has a time period $$ T_{1} $$. When the lift moves downward with constant velocity, the time period is $$ T_{2} $$, then
  • $$

    T_{2}

    $$ is infinity
  • $$

    T_{2}

    $$>$$

    T_{1}

    $$
  • $$

    T_{2}

    $$<$$

    T_{1}

    $$
  • $$

    T_{2}

    $$=$$

    T_{1}

    $$
If velocity of a particle in SHM at x=4 m and x=5 m are 15 m/s and 13 m/s then its time period will be:
  • $$ \pi $$ /4
  • $$ \pi $$ /2
  • $$ \pi $$
  • 4$$ \pi $$ /5
The time period of mass suspended from a spring is T. If the spring is cut into four equal parts and the same mass is suspended from one of the parts, then the new time period will be:
  • $$\cfrac T 4$$
  • $$T$$
  • $$\cfrac T 2$$
  • $$2 T $$
A body oscillates with SHM according to the equation $$x=(5.0\quad m)\cos { [(2\pi \quad rad\quad { s }^{ -1 } } )t+{ \pi  }/{ 4] }$$
At t=1.5 s, its acceleration is:
  • $$-139.56\quad { m }/{ { s }^{ 2 } }$$
  • $$139.56\quad { m }/{ { s }^{ 2 } }$$
  • $$69.78\quad { m }/{ { s }^{ 2 } }$$
  • $$-69.78\quad { m }/{ { s }^{ 2 } }$$
A coin is placed on a horizontal platform, which undergoes horizontal simple harmonic motion about a mean position $$O$$.The coin does not slip on the platform. The force of friction acting on the coin is $$F$$.
  • $$F$$ is always directed towards $$O$$
  • $$F$$ is directed towards $$O$$ when the coin is moving away from $$O$$, and away from $$O$$ when the coin moves towards $$O$$
  • $$F=0$$ when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.
  • $$F$$ is maximum when the coin and platform come to rest momentarily at the extreme position of the harmonic motion.
A body attached to the lower end of a vertical spring oscillates with time period of 1 sec. The time period when two such springs are connected one below another is approximately:
  • 0.7 sec
  • 1 sec
  • 1.4 sec
  • 2 sec
When the displacement is half of the amplitude, then what fraction of total energy of a simple harmonic oscillator is kinetic:-
  • $$\dfrac {3}{4}th$$
  • $$\dfrac {2}{7}th$$
  • $$\dfrac {5}{7}th$$
  • $$\dfrac {2}{8}th$$
a simple harmonic has an amplitude A and time period T. Find the time required by it to travel directly from $$x=0$$ to $$\, x=\frac { A}{ \sqrt { 2} } $$
  • T
  • 2T
  • T/2
  • T/8
Two bodies A and B of equal mass are suspended from separate massless springs of spring constant $$k_1$$ and $$k_2$$ respectively. If the bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of A to that of B is
  • $$k_1/k_2$$
  • $$\sqrt{k_1/k_2}$$
  • $$k_2/k_1$$
  • $$\sqrt{k_2/k_1}$$
A simple harmonic wave is represented by the relation $$y\left ( x,t \right )=a_{0}sin2\pi \left ( vt-\dfrac{x}{\lambda } \right )$$ if the maximum particle velocity is three times the wave velocity, te wavelength $$\lambda $$ of the wave is-
  • $$\pi a_{0}/3$$
  • $$2\pi a_{0}/3$$
  • $$\pi a_{0}$$
  • $$\pi a_{0}/2$$
An object undergoing SHM taken 0.5 s to travel from one point of zero velocity to the next such point. The distance between those point is 50 cm. The period, frequency and amplitude of the motion is:
  • $$1s, 1Hz,25 cm$$
  • $$2s.1Hz,50 cm$$
  • $$1s, 2Hz,25 cm$$
  • $$2s, 2Hz,50 cm$$
A particle is executing SHM with time period T Starting from mean position, time taken by it to complete $$\dfrac { 5 }{ 8 } $$ oscillations is:
  • $$\dfrac { T }{ 12 } $$
  • $$\dfrac { T }{ 6 } $$
  • $$\dfrac { 5T }{ 12 } $$
  • $$\dfrac { 7T }{ 12 } $$
The equation of motion of a particle executing SHM is $$\left( \dfrac { d ^ { 2 } x } { d t ^ { 2 } } \right) + k x = 0$$ The time period of the particle will be 
  • $$\dfrac{2 \pi }{ \sqrt { k }}$$
  • $$\dfrac{2 \pi }{ \mathrm { k }}$$
  • $$2 \pi k$$
  • $$2 \pi \sqrt { k }$$
The function $$sin^{2}(\omega t)$$  represents : 
  • a simple harmonic motion with a period $$2\pi /\omega $$.
  • a simple harmonic motion with a period $$\pi /\omega $$
  • a periodic, but not simple harmonic motion with period $$2\pi /\omega $$
  • a periodic, but not simple harmonic motion with period $$\pi /\omega $$
A particle performs $$SHM$$ on x-axis with amplitude $$A$$ and time period $$T$$. The time taken by the particle to a distance $$A/5$$ starting from rest is:
  • $$\dfrac { T }{ 20 }$$
  • $$\dfrac { T }{ 2\pi } \cos^{-1} \left(\dfrac { 4 }{ 5 }\right)$$
  • $$\dfrac { T }{ 2\pi } \cos^{-1} \left(\dfrac { 1 }{ 5 }\right)$$
  • $$\dfrac { T }{ 2\pi } \sin^{-1} \left(\dfrac { 4 }{ 5 }\right)$$

The figure shows the displacement time graph of a particle executing S.H.M.
If the time period of oscillation is $$2 s$$ the equation of motion of its SHM 


1391260_dbb768ce8618453a8f6dbba8acc234af.png
  • $$
    x = 10 \sin ( \pi t + \pi / 3 )
    $$
  • $$
    x = 10 \sin \pi t
    $$
  • $$
    x = 10 \sin ( \pi t + \pi / 6 )
    $$
  • $$
    x = 10 \sin (2 \pi t + \pi / 6 )
    $$
The displacement $$x$$ (in metres) of a particle performing simple harmonic motion is related to time $$t$$ (in seconds as} $$x = 0.05 \cos \left( 4 \pi t + \frac { \pi } { 4 } \right)$$. The frequency of the motion will be
  • $$0.5$$$$\mathrm { Hz }$$
  • $$1.0$$$$\mathrm { Hz }$$
  • $$1.5$$$$\mathrm { Hz }$$
  • $$2.0$$$$\mathrm { Hz }$$

Restoring force on the bob of a simple pendulum of mass 100 gm when its amplitude is $${1^\circ}$$ :
  • $$1.7 N$$
  • $$0.17 N$$
  • $$0.017 N$$
  • $$0.034 N$$
A particle of mass $$0.1 kg$$ executes SHM under a force $$F = (-10 x) N$$. Speed of particle at mean position is $$6 m/s$$. Then amplitude of oscillations is 
  • $$0.6 m$$
  • $$0.2 m$$
  • $$0.4 m$$
  • $$0.1 m$$
0:0:1


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