Tag: introduction to sound

Questions Related to introduction to sound

A simple harmonic oscillator of angular frequency $2\ rad\ s^{-1}$ 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

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

  1. $\sin t + \dfrac {1}{2} \sin 2t$

  2. $\sin t + \dfrac {1}{2} \cos 2t$

  3. $\cos t - \dfrac {1}{2} \sin 2t$

  4. $\sin t - \dfrac {1}{2} \sin 2t$

Show answer
Correct Option: D

A body of mass $\text{600 gm}$ is attached to a spring of spring constant $\text{k = 100 N/m}$ and it is performing damped oscillations.  If damping constant is $0.2$ and driving force is $F = F _{0}$  $cos(\omega t)$  where $F _{0}=20N$  Find the amplitude of oscillation at resonance. 

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

  1. $\text{4.1 m}$

  2. $\text{0.57 m}$

  3. $\text{7.7 m}$

  4. $\text{0.98 m}$

Show answer
Correct Option: C
Explanation:

As we know that the amplitude of forced oscillation is given as

$A=\dfrac{F _0}{\sqrt{m^2(\omega^2-\omega _d^2)^2+\omega _d^2b^2}}$

Here we know that when oscillator is in resonance then,

$\omega=\omega _d$

so we have

$A=\dfrac{F _0}{\omega _d b}$

$F _0=20\,N$

$m = 600\, g$

$\omega=\sqrt{\dfrac km}$

$\omega=\sqrt{\dfrac{100}{0.6}}$

$\omega=12.9\,rad/sec$

Now we have

$A=\dfrac{20}{12.9\times 0.2}$

$A=7.7\,m$

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas $ \mathrm{V} _{0}  $ and its pressure is $  \mathrm{P} _{0} $ The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency.

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

  1. $ \dfrac{1}{2 \pi} \dfrac{\mathrm{A} \gamma P _{0}}{V _{0} M} $

  2. $ \dfrac{1}{2 \pi} \dfrac{V _{0} M P _{0}}{A^{2} \gamma} $

  3. $ \dfrac{1}{2 \pi} \sqrt{\dfrac{A^{2} \gamma P _{0}}{M V _{0}}} $

  4. $ \dfrac{1}{2 \pi} \sqrt{\dfrac{M V _{0}}{A \gamma P _{0}}} $

Show answer
Correct Option: C

The amplitude of a damped oscillator becomes half on one minute. The amplitude after 3 minute will be $\displaystyle\dfrac{1}{X}$ times the original, where $X$ is

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

  1. $2\times 3$

  2. $2^3$

  3. $3^2$

  4. $3\times 2^2$

Show answer
Correct Option: B

The equation of a damped simple harmonic motion is $ m \frac {d^2x}{dt^2} + b \frac {dx}{dt} + kx=0 . $ Then the angular frequency of oscillation is:

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

  1. $ \omega = ( \frac {k}{m}+\frac {b}{4m})^{1/2} $

  2. $ \omega = ( \frac {k}{m}-\frac {b}{4m})^{1/2} $

  3. $ \omega = ( \frac {k}{m}+\frac {b^2}{4m})^{1/2} $

  4. $ \omega = ( \frac {k}{m}-\frac {b^2}{4m^2})^{1/2} $

Show answer
Correct Option: D

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

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

  1. $0.7$

  2. $0.81$

  3. $0.729$

  4. $0.6$

Show answer
Correct Option: C

A lightly damped oscillator with a frequency $\left( \omega  \right) $  is set in motion by harmonic driving force of frequency $\left( n \right) $. When $n\ll \omega $, then response of the oscillator is controlled by

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

  1. Oscillator frequency

  2. spring constant

  3. Damping coefficient

  4. Inertia of the mass

Show answer
Correct Option: B,C

On account of damping , the frequency of a vibrating body

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

  1. remains unaffceted

  2. increases

  3. decreases

  4. changes erratically

Show answer
Correct Option: C
Explanation:

Damping is caused by opposing force, which decreases the frequency.

Ans: C

In damped oscillations, the amplitude after $50$ oscillations is $0.8\;a _0$, where $a _0$ is the initial amplitude, then the amplitude after $150$ oscillations is

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

  1. $0.512\;a _0$

  2. $0.280\;a _0$

  3. Zero

  4. $a _0$

Show answer
Correct Option: A
Explanation:

The amplitude, a, at time $t$ is given by $a=a _0\;exp(-\,\alpha t)$



$a _{50}=a _0\;exp(-\alpha\times 50T)=0.80\;a _0$



where $T$ is the period of oscillation



$a _{150}=a _0\;exp(-a\times 150T)$



$=a _0\;(0.8)^3=0.512\,a _0$

When an oscillator completes $100$ oscillations its amplitude reduces to $\displaystyle\dfrac{1}{3}$ of its initial value. What will be its amplitude when it completes $200$ oscillations?

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

  1. $\displaystyle\dfrac{1}{8}$

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

  3. $\displaystyle\dfrac{1}{6}$

  4. $\displaystyle\dfrac{1}{9}$

Show answer
Correct Option: D
Explanation:


Its is a damped oscillation, where amplitude of oscillation at time $t$ is given by $A = a _0e^{-\gamma t}$
where $a _0 = $ initial amplitude of oscillation
$\quad \gamma = $ damping constant
As per question, $\displaystyle\dfrac{a _0}{3} = a _0e^{-\gamma100/v}\quad                    ...(i)$
(where $v$ is the frequency of oscillation)
and $A = a _0e^{-\gamma200/v} \quad                ...(ii)$
From $(i)$; $\quad \displaystyle\dfrac{a _0}{3} = a _0e^{-\gamma\times100/v} \quad            ...(iii)$
Dividing equation $(ii)$ by $(iii)$, we have

$\quad \displaystyle\dfrac{A}{a _0(1/3)} = \displaystyle\dfrac{e^{-\gamma\times200/v}}{e^{-\gamma\times100/v}} = e^{-\gamma\times100/v} = \displaystyle\dfrac{1}{3}$

or $A = a _0\times\displaystyle\dfrac{1}{3}\times\displaystyle\dfrac{1}{3} = \displaystyle\dfrac{1}{9}a _0$