Tag: resonance

Quizzes (0) Questions (87)

Questions Related to resonance

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


Multiple choice physics option b: engineering physics introduction to sound free, forced and damped oscillations resonance

In reality, a spring won't oscillate for ever.               will                the amplitude of oscillation until eventually the system is at rest.

  1. Frictional force, increase

  2. Viscous force, decrease

  3. Frictional force, decrease

  4. Viscous force, increase
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In reality, a spring won't oscillate forever. Frictional force will decrease the oscillation until eventually, the system is at rest.