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

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

  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. $\sin t - \dfrac {1}{2} \sin 2t$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a forced oscillation problem. The steady-state solution for x(t) with a driving force F sin(t) involves terms of sin(t) and sin(omega_0 * t). Given omega_0 = 2, the solution is a combination of the driving frequency and natural frequency.

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

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. 

  1. $\text{4.1 m}$
  2. $\text{0.57 m}$
  3. $\text{7.7 m}$
  4. $\text{0.98 m}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

$A=7.7\,m$

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

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.

  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}}} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an adiabatic process, PV^gamma = constant. The restoring force for a small displacement x is F = -A * dP = -A * (gamma * P0 / V0) * (A * x). This leads to the SHM equation with omega^2 = (A^2 * gamma * P0) / (M * V0). Frequency f = omega / (2 * pi).

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

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

  1. $2\times 3$
  2. $2^3$
  3. $3^2$
  4. $3\times 2^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A(t) = A0 * exp(-kt). A(1) = A0/2, so exp(-k) = 1/2. A(3) = A0 * (exp(-k))^3 = A0 * (1/2)^3 = A0/8. Thus X = 8 = 2^3.

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

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:

  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} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The equation of motion is m*x'' + b*x' + k*x = 0. The angular frequency of the damped oscillation is omega = sqrt(k/m - (b/2m)^2) = sqrt(k/m - b^2/(4m^2)).

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

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.

  1. $0.7$
  2. $0.81$
  3. $0.729$
  4. $0.6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A(5) = 0.9 * A0. A(15) = A0 * (exp(-5k))^3 = A0 * (0.9)^3 = 0.729 * A0. Thus alpha = 0.729.

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

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

  1. $0.512\;a _0$
  2. $0.280\;a _0$

  3. Zero
  4. $a _0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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



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



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

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

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?

  1. $\displaystyle\dfrac{1}{8}$
  2. $\displaystyle\dfrac{2}{3}$
  3. $\displaystyle\dfrac{1}{6}$
  4. $\displaystyle\dfrac{1}{9}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $A = a _0\times\displaystyle\dfrac{1}{3}\times\displaystyle\dfrac{1}{3} = \displaystyle\dfrac{1}{9}a _0$