Tag: oscillations

Questions Related to oscillations

What impulse need to be given to a body of mass $m$, released from the surface of earth along a straight tunnel passsing through centre of earth, at the centre of earth, to bring it to rest(Mass of earth $M$, radius of earth R) 

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $m \sqrt { \dfrac { G M } { R } }$

  2. $\sqrt { \dfrac { G M m } { R } }$

  3. $m \sqrt { \dfrac { G M } {2 R } }$

  4. $zero$

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Correct Option: D

A particle is suspended from a light vertical inelastic string of length 'l' from a fixed support. At its equilibrium position, it is projected horizontally with a speed $\sqrt{6gl}$. Find the ratio of tension on string, its horizontal position to that in vertically above the point of support.

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $2:1$

  2. $4:1$

  3. $3:1$

  4. $5:1$

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Correct Option: B

The amplitude of a damped harmonic oscillator becomes halved in $\ minute$. After three minutes, the amplitude will becomes $\dfrac{1}{x}$ of initial amplitude, where $x$ is ?

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $8$

  2. $2$

  3. $3$

  4. $4$

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Correct Option: A

A particle performing SHM is found at its equilibrium at $  t=1\ sec$ and it is found to have a speed of $0.25 \mathrm{m} / \mathrm{s}  $ at $  \mathrm{t}=2\ \mathrm{sec}  $ . If the period of oscillation is $6\ \mathrm{sec}  $. Calculate amplitude of oscillation

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $ \frac{3}{2 \pi} \mathrm{m} $

  2. $ \frac{3}{ \pi} \mathrm{m} $

  3. $ \frac{6}{2 \pi} \mathrm{m} $

  4. $ \frac{6}{ \pi} \mathrm{m} $

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Correct Option: A

A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F \ sin \omega.$ If the amplitude of the particle is maximum for $\omega = \omega _1$ and the energy of the particle is maximum for $\omega = \omega _2$ then (where $\omega _0$ natural frequency of oscillation of particle)

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $\omega _1 = \omega _0 \ and \ \omega _2 \neq \omega _0$

  2. $\omega _1 = \omega _0 \ and \ \omega _2 = \omega _0$

  3. $\omega _1 \neq \omega _0 \ and \ \omega _2 =\omega _0$

  4. $\omega _1 \neq \omega _0 \ and \ \omega _2 \neq \omega _0$

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Correct Option: C
Explanation:

As we know the energy of the particle is maximum at natural frequency. Since, the restoring force is proportional to displacement and resisting force is proportional to velocity. So the correct option is ${{\omega } _{0}}={{\omega } _{2}}\,\And \,{{\omega } _{1}}\ne\,{{\omega } _{0}}$

Few particles undergo damped harmonic motion. Values for the spring constant $k$ , the damping constant $b$ , and the mass $m$ are given below. Which leads to the smallest rate of loss of mechanical energy at the initial moment?

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $ k = 100N/m , m = 50 g, b = 8 g/s $

  2. $ k = 150 N/m , m = 50 g, b = 5 g/s $

  3. $ k = 150N/m , m = 10g, b = 8 g/s $

  4. $ k = 200N/m , m = 8g, b = 6 g/s $

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Correct Option: B

A bar magnet oscillates with a frequency of$ 10 $ oscillations per minute. When another bar magnet is placed on its axis at a small distance, it oscillates at $14$ oscillations per minute. Now, the second bar magnet is turned so that poles are instantaneous, keeping the location same. The new frequency of oscillation will be 

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $2$ vibrations/min

  2. $4$ vibrations/min

  3. $10$ vibrations/min

  4. $14$ vibrations/min

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Correct Option: A
Explanation:

$\dfrac{60}{10}= 2\pi \sqrt{\dfrac{l}{MB _H}}$
$\dfrac{60}{14}= 2\pi \sqrt{\dfrac{l}{M(MB _H)}}$
$\therefore \dfrac{7}{5} = \sqrt{B _H +B}{B _H} $ or $ B= \dfrac{24}{25}B _H$
Hence,
$\dfrac{60}{10}= 2\pi \sqrt{\dfrac{l}{M(B _H-B)}}= 2\pi \sqrt{\dfrac{l}{MB(1-24/25)}}$
$= 5\times 2\pi \sqrt{\dfrac{l}{2MB}} = 5 \times \dfrac{60}{10}= 30$
$\therefore f= \dfrac{60}{30} = 2$ vibrations/ min

The angular frequency of the damped oscillator is given by $\omega =\sqrt{\left(\frac{k}{m} -\dfrac{r^2}{4m^2}\right)}$ 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 $8%$, the changed in time period compared to the undamped oscillator is approximately as follows:  

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. Increases by 1%

  2. Decreases by 1%

  3. Decreases by 8%

  4. increases by 8%

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Correct Option: B
Explanation:

$\omega = \sqrt {\dfrac{k}{m}-\dfrac{r^2}{4m^2}}= \sqrt{\dfrac{k}{m}}\sqrt{1-\dfrac{r^2}{4mk}}$
 $\approx \omega _o \left(1-\dfrac{r^2}{8mk}\right) \approx$ (1 - 1%)

The amplitude of a damped oscillator becomes $\left (\dfrac {1}{3}\right )rd$ in $2s$. If its amplitude after $6\ s$ in $\dfrac {1}{n}$ times the original amplitude, the value of $n$ is

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $3^{2}$

  2. $3\sqrt {2}$

  3. $3^{3}$

  4. $2^{3}$

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Correct Option: C
Explanation:

Let original amplitude $=A$

Amplitude after 2 sec=$\dfrac{A}{3}$
Amplitude after next 2 sec=$\dfrac{A}{3}\times \dfrac{1}{3}=\dfrac{A}{9}$
Amplitude again  after 2 sec=$\dfrac{1}{3}\times \dfrac{A}{9}=\dfrac{A}{27}=\dfrac{A}{3^3}$
Here $n=3^3$

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be (intial amplitude =$A _{0}$)

physics oscillations introduction to sound free, forced and damped oscillations resonance

  1. $\dfrac{1}{4}A _{0}$

  2. $\dfrac{1}{2}A _{0}$

  3. $\dfrac{1}{5}A _{0}$

  4. $\dfrac{1}{7}A _{0}$

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Correct Option: A
Explanation:

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be

 

Amplitude is given by:

$A={{A} _{o}}{{e}^{-\alpha t}}$

Where, $A$ is amplitude at time t.

t is time

${{A} _{0}}$ is initial aplitude

$\alpha $ is constant

At t = 1s

$A=\dfrac{{{A} _{0}}}{2}$

So,

$ \dfrac{{{A} _{0}}}{2}={{A} _{0}}{{e}^{-\alpha }} $

$ {{e}^{-\alpha }}=\dfrac{1}{2} $

At t = 2s

$ A={{A} _{0}}{{e}^{-2\alpha }} $

$ A={{A} _{0}}{{(\dfrac{1}{2})}^{2}} $

$ A=\dfrac{{{A} _{0}}}{4} $