Tag: free, forced and damped oscillations

Questions Related to free, forced and damped oscillations

The length and diameter of a metal wire is doubled. The fundamental frequency of vibration will change from '$n$' to (Tension being kept constant and material of both the wires is same)

physics free, damped and forced oscillations melde's experiment free, forced and damped oscillations sonometer and laws of transverse vibrations

  1. $\dfrac { n }{ 4 } $

  2. $\dfrac { n }{ 8 } $

  3. $\dfrac { n }{ 12 } $

  4. $\dfrac { n }{ 16 } $

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Correct Option: A
Explanation:

Fundamental frequency of vibration $n = \dfrac{v}{2L} \sqrt{\dfrac{T}{\mu}}$ 

where $\mu$ is the mass per unit length of the wire i.e. $\mu = \dfrac{M}{L}$
Mass of the wire $M = \rho (\dfrac{4\pi}{3} R^3)$
$\implies$ $n = \dfrac{v}{2L} .\sqrt{\dfrac{TL}{\rho \dfrac{4\pi }{3} R^3}}$
$\implies$ $n \propto \dfrac{1}{R\sqrt{LR}}$      .....(1)
Given :  $L _2 = 2L$  $R _2 = 2R$
From equation (1), we get  $\dfrac{n _2}{n} = \dfrac{R \sqrt{RL}}{R _2 \sqrt{L _2 R _2}}$
Or  $\dfrac{n _2}{n} = \dfrac{R \sqrt{R L}}{(2R) \sqrt{(2L) (2R)}}   = \dfrac{1}{4}$
$\implies$  $n _2 = \dfrac{n}{4}$

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:

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  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. $\omega _1< \omega _2$

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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

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $f$

  2. $2\ f$

  3. $f/3$

  4. $3\ f$

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Correct Option: C

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

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

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

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

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

  4. All the above are true

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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: 

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. $2\pi$

  2. $\pi$

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

  4. $\sqrt {2}\pi$

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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  

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  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) } $

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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 :  

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  1. 0.7

  2. 0.81

  3. 0.729

  4. 0.6

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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.

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

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

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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?

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

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

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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:

physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

  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} } $

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Correct Option: B