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Forced vibrations - class-XI

Description: forced vibrations
Number of Questions: 29
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Tags: physics oscillatory motion oscillation and waves option b: engineering physics free, damped and forced oscillations oscillations
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In forced oscillation of a particle the amplitude is maximum for a frequency $\omega _1$ of force, while the energy is maximum for a frequency $\omega _2$ of the force, then:

  1. $\omega _1= \omega _2$

  2. $\omega _1> \omega _2$

  3. $\omega _1 < \omega _2$ when damping is small and $\omega _1> \omega _2$ when damping is large

  4. $\omega _1< \omega _2$


Correct Option: A
Explanation:

For the amplitude of oscillation and energy to be maximum, the frequency of force must be equal to the initial frequency and this is only possible in resonance. In resonance state $ \omega _1 = \omega _2$.

A weightless spring has a force constant $k$ oscillates  with frequency $f$ when a mass $m$ is suspended from it. The spring is cut into three equal parts and a mass $3\ m$ is suspended from it. The frequency of oscillation of one part  will now becomes

  1. $f$

  2. $2\ f$

  3. $f/3$

  4. $3\ f$


Correct Option: C

If density (D) acceleration (a) and force (F) are taken as basic quantities,then Time period has dimensions

  1. $\dfrac {1} {6}$ in F

  2. $-\dfrac {1} {6}$ in F

  3. $-\dfrac {2} {3}$ in F

  4. All the above are true


Correct Option: A
Explanation:

The density has dimension ML-3

so take that as D

the acceleration has dimension LT-2

take it as A

the force has dimension MLT-2

Take it as F

so time period has dimension as D-1/6 A -2/3 F1/6

The potential energy of a particle of mass $1\ kg$ in motion along the $x-$axis is given by: $U=4(1-\cos 2x)\ l$, where $x$ is in metres. The period of small oscillations (in sec) is: 

  1. $2\pi$

  2. $\pi$

  3. $\dfrac{\pi}{2}$

  4. $\sqrt {2}\pi$


Correct Option: C

A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement is kept on a horizontal table (the surface of concave mirror is frictionless and sliding not rolling). If the sphere is displaced from its equilibrium position and left, then it executes S.H.M. The period of oscillation will be  

  1. $\pi \times { \left( \dfrac { (R-r)1.4 }{ g } \right) } $

  2. $2\pi \times  { \left( \dfrac { R-r }{ g } \right) } $

  3. $\sqrt [ 2\pi ]{ \left( \dfrac { r\quad R }{ g } \right) } $

  4. ${ \left( \dfrac { R }{ g\quad r } \right) } $


Correct Option: B

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5 s .In another 10 s it will decrease to $\alpha $ times its original magnitude where $\alpha $ equals :  

  1. 0.7

  2. 0.81

  3. 0.729

  4. 0.6


Correct Option: C

Three infinitely long thin wires, each carrying current I in the same direction, are in the $x-y$ plane of a gravity free space. The central wire is along the y-axis while the other two are along $x=\pm\ d$.
(a) Find the locus of the points for which the magnetic field $B$ is zero.
(b) If the central wire is displaced along the z-direction by a small amount and released, show that it will execute simple harmonic motion. If the linear density of the wires is $\lambda$, find the frequency of oscillation.

  1. $\dfrac{1}{4\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.

  2. $\dfrac{1}{2\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.

  3. $\dfrac{2}{2\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.

  4. $\dfrac{3}{2\pi}\sqrt{\dfrac{\mu _oI^2}{\pi\lambda d^2}}$.


Correct Option: B

A student performs an experiment for determination of $\Bigg \lgroup g = \frac{4\pi^2 l}{T^2} \Bigg \rgroup$, l = 1m, and he commits an error of $\Delta l$ For T he takes the time of n oscillations with the stop watch of least count $\Delta T$ and he commits a human error of 0.1 s. For which of the following data, the measurement of g will be most accurate?

  1. $\Delta L = 0.5, \ \Delta T = 0.1 , \ n = 20$

  2. $\Delta L = 0.5, \ \Delta T = 0.1 , \ n = 50$

  3. $\Delta L = 0.5, \ \Delta T = 0.01 , \ n = 20$

  4. $\Delta L = 0.5, \ \Delta T = 0.05 , \ n = 50$


Correct Option: B
Explanation:

It is given that $g$ is determined by $:g = \frac{{4{\pi ^2}l}}{{{T^2}}}$

Taking log and differentiating$,$
$ \Rightarrow \frac{{\Delta g}}{g} = \frac{{\Delta l}}{l} + \frac{{2\Delta T}}{T}$ 
Now$,$ $g$ is most accurate in the case in which error in $g\left( {l.e\,\Delta g} \right)$ is minimum$.$ In case $B,$ number of repetitions to perform the experiment is maximum and ${\Delta g}$  is minimum$.$
Hence,
option $(B)$ is correct answer.

A particle moves such that its acceleration is given by : $\alpha=-\beta(x-2)$
Here :$\beta$ is a positive constant and x the position from oigin. Time period of oscillations is:

  1. $2\pi+\sqrt\beta$

  2. $2\pi +\sqrt { \cfrac { 1 }{ \beta } } $

  3. $2\pi+\sqrt{\beta+2}$

  4. $2\pi +\sqrt { \cfrac { 1 }{ \beta +2} } $


Correct Option: B

A simple pendulum suspended from the ceiling of a stationary trolley has a length $l$ its period of oscillation is $2\pi\sqrt{l/g}$. Whqat will be its period of oscillation if the trolley moves forward with an acceleration $f$?

  1. g

  2. l

  3. d

  4. m


Correct Option: A

Find the time period of small oscillations of the following systems. 

  1. A metre stick suspended through the 20 cm mark.

  2. A ring of mass m and radius r suspended through a point on its perphery.

  3. A uniform square plate of edge a suspended through a corner.

  4. A uniform disc of mass m and radius r suspended through a point r/2 away from the centre.


Correct Option: A

A uniform circular disc of radius $R$ oscillates about a horizontal axis in its own plane. The distance of the axis from the center for the period of oscillation is maximum, will be :

  1. $R$

  2. $\dfrac{R}{\sqrt 2}$

  3. $\dfrac{R}{3}$

  4. $\dfrac{R}{4}$


Correct Option: B

Find the frequency of oscillation of the spheres

  1. $\frac{1}{{2\pi }}\sqrt {\dfrac{{35K}}{{46m}}} $

  2. $\frac{1}{{2\pi }}\sqrt {\dfrac{{46K}}{{35m}}} $

  3. $\frac{1}{{2\pi }}\sqrt {\dfrac{{25K}}{{46m}}} $

  4. $\frac{1}{{2\pi }}\sqrt {\dfrac{{21K}}{{46m}}} $


Correct Option: A

Find the frequency of oscillation of the spheres

  1. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 35K }{ 46m } } $

  2. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 46K }{ 35m } } $

  3. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 25K }{ 46m } } $

  4. $ \dfrac { 1 }{ 2\pi } \sqrt { \dfrac {21K }{ 46m } } $


Correct Option: C

A particle of mass m is in one dimensional potential field and its potential energy is given by the following equation U(x)=${U _0}\left( {1 - \cos \;aX} \right)\;where\;{U _0}$ and $\alpha $ constants.The period of the particle for small oscillations near the equilibrium will be-

  1. $2\pi \sqrt {\dfrac{{m{\alpha ^2}}}{{m{\alpha ^2}{U _0}}}} $

  2. $2\pi \sqrt {m{\alpha ^2}{U _0}} $

  3. $2\pi \sqrt {\dfrac{m}{{{\alpha ^2}{U _0}}}} $

  4. $2\pi \sqrt {\dfrac{{{\alpha ^2}{U _0}}}{m}} $


Correct Option: A

A boy is playing on a swing in sitting position. the time period of oscillation of the swing is T, if the boy stands up, the time period of oscillation of the spring will be:

  1. $T$

  2. $ Less \ than T$

  3. $More\  than T$

  4. such as cannot be predicted


Correct Option: A

The time taken by a particle performing S.H.M. to pass from point $ A  $ to $  B  $ where its velocities are same is $2$ seconds. After another 2 seconds it returns to $ \mathrm{B}  $ . The time period of oscillation is (in seconds):

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: D
Explanation:

According to the question, a and b points are such that they are

 located at same distances from the equilibrium position. 
Here also it is said that velocity is same 
,i.e; not only the magnitude but also the directions are same.
 So, total time period of oscillation $= 2×$( time taken to go from a to b
 + the next time taken to return at b) $= 2×(2+2)= 8$ sec.

A student measures the time period of oscillation of a simple pendulum. He uses the data to estimate the acceleration due to gravity 9g) at that place. If the maximum percentage error in measurement of length pendulum and that in time are $ e _{1} $ and $ e _{2} $ respectively then percentage error estimation of ''g'' is :

  1. $

    e _{1}+2 e _{2}

    $

  2. $

    2 e 1+e 2

    $

  3. $

    e 1+e _{2}

    $

  4. $

    e 1-e _{2}

    $


Correct Option: A

The phase of particle in SHM is found to increase by $14 \pi$ in 3.5 sec. Its frequency of oscillation is

  1. $2 Hz$

  2. $1/2 Hz$

  3. $1 Hz$

  4. $2 \pi Hz$


Correct Option: B

Frequency of oscillation of a body is $6\;Hz$ when force $F _1$ is applied and $8\;Hz$ when $F _2$ is applied. If both forces $F _1\;&amp;\;F _2$ are applied together then, the frequency of oscillation is :

  1. $14\;Hz$

  2. $2\;Hz$

  3. $10\;Hz$

  4. $10\surd{2}\;Hz$


Correct Option: C
Explanation:

According to question,

$F _1=-K _1\,x\;\&\;F _2=-K _2\,x$

So $n _1=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1}{m}}=6\;Hz;$

$n _2=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _2}{m}}=8\;Hz$

Now $F=F _1+F _2=-(K _1+K _2)x$

Therefore $n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{K _1+K _2}{m}}$

$\Rightarrow n=\displaystyle\frac{1}{2\pi}\sqrt{\displaystyle\frac{4\pi^2\,n^2 _1\,m+4\pi^2\,n^2 _2\,m}{m}}$$=\sqrt{n _1^2+n _2^2}=\sqrt{8^2+6^2}$

$=10\;Hz$

The angular frequency of the damped oscillator is given by $\omega =\sqrt { \left( \dfrac { k }{ m } -\dfrac { { r }^{ 2 } }{ 4{ m }^{ 2 } }  \right)  }$ , where k is the spring constant, $m$ is the mass of the oscillator and $r$ is the damping constant. If the ratio $\dfrac { { r }^{ 2 } }{ mk }$ is $80$%, the change in time period compared to the undamped oscillator is approximately as follows:

  1. Decreases by $1$%

  2. Increases by $8$%

  3. Increases by $1$%

  4. Decreases by $8$%


Correct Option: A

In case of a forced vibration, the resonance wave becomes very sharp when the :

  1. Damping force is small

  2. Restoring force is small

  3. Applied periodic force is small

  4. Quality factor is small


Correct Option: A
Explanation:

In forced vibration, the resonance wave becomes very sharp when damping force is small (i.e. negligible)

The potential energy of a particle of mass 1 kg in motion along the x-axis is given by U = 4(1 - cos2x) J. Here x is in meter. The period of small oscillations (in sec) is _______.

  1. $2\pi$

  2. $\pi$

  3. $\dfrac{\pi}{2}$

  4. $\sqrt{2\pi}$


Correct Option: C
Explanation:
$M=1\ kg$
$U=u(1-\cos^2 x)$
OR $U=8\sin^2 x\quad [\because \ 1-\cos \theta =2\sin^2 \theta]$
$f=-\dfrac {du}{dx}$
$f=-u [0-(-\sin 2x) (2)]$
$f=-8\sin 2x\ N$
$a=-8\sin 2x \ m/ \sec^2 \quad [\because \ m=1\ kg]$
$\Rightarrow \ a\simeq -8(2x)$
$a=-16x$ [For small $x, \sin x \approx x$]
$\Rightarrow \ w^2 =4^2$
$w=4\ $ rad /ac
$T=2\dfrac {\pi}{w}=\dfrac {\pi}{2}\sec$

The period of oscillation of a simple pendulum of constant length is independent of

  1. size of the bob

  2. shape of the bob

  3. mass of bob

  4. all of these


Correct Option: D
Explanation:

$T=2 \pi \sqrt{\dfrac{L}{g}}$               (where L= length of sting)
From above equation, Time period only depend on the length of the string and g.

Option d 

Assertion : In damped oscillations, the energy of the system is dissipated continuously.
Reason : For the small damping, the oscillations remain approximately periodic.

  1. If both assertion and reason are true and reason is the correct explanation of assertion.

  2. If both assertion and reason are true and reason is not the correct explanation of assertion.

  3. If assertion is true but reason is false.

  4. If both assertion and reason are false.


Correct Option: B
Explanation:

In damped oscillation like the motion of simple pendulum swinging in air, motion dies out eventually. This is because of the air drag and the friction at the support oppose the motion of the pendulum and dissipate its energy gradually.

Assertion : In forced oscillations, the steady state motion of the particle is simple harmonic.
Reason : Then the frequency of particle after the free oscillations die out, is the natural frequency of the particle.

  1. If both assertion and reason are true and reason is the correct explanation of assertion.

  2. If both assertion and reason are true and reason is not the correct explanation of assertion.

  3. If assertion is true but reason is false.

  4. If both assertion and reason are false.


Correct Option: C
Explanation:

In forced oscillations, the frequency of particle after free oscillations die out, is the frequency of the driving force, not the natural frequency of the particle.

A force F= -4x-8 is acting on a block where x is position of block in meter. The energy of oscillation is 32 J, the block oscillate between two points Position of extreme position is:

  1. 6

  2. 0

  3. 4

  4. 3


Correct Option: A
Explanation:

$F=-4x-8\F=-4(x+2)\ \Rightarrow F=-4[x-(-2)]\F=-kx\rightarrow$ Distance mean position. So, this will be SHM

with $x=-2$ as mean position 
as $K _SHM=4$
Energy of SHM$=\cfrac{1}{2}KA^2=32\ \Rightarrow A^2=\cfrac{2\times32}{K}=\cfrac{2\times32}{4}=16\Rightarrow A=4$
and mean position $=-2$
Extereme position $=(-2+4)\quad and (-2-4)\=2\quad and\quad -6$

A block of $0.5\ kg$ is placed on a horizontal platform. The system is making vertical oscillations about a fixed point with a frequency of $0.5\ Hz.$ Find the maximum amplitude of oscillation if the block is not to lose contact with the horizontal platform?

  1. $0.6542\ m$

  2. $0.9927\ m$

  3. $0.7428\ m$

  4. $0.852\ m$


Correct Option: D

A highly rigid cubical block A of small mass M and side L is fixed rigidly on another cubical block B of the same dimensions and of low modulus of rigidity $\eta $ such that the lower face of A completely covers the upper face of B.  The lower face of B is rigidly held on horizontal surface.  A small force is applied perpendicular to the side faces of A.  After the force is withdrawn, block A executes small oscillations the time period of which is given by 

  1. $2\pi \sqrt{M\eta L}$

  2. $2\pi \sqrt{\frac{M-\eta }{L}}$

  3. $2\pi \sqrt{\frac{M-L}{\eta }}$

  4. $2\pi \sqrt{\frac{M-N}{\eta L}}$


Correct Option: A
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