Tag: oscillations

Quizzes (16) Questions (206)

Questions Related to oscillations

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 oscillations introduction to sound free, forced and damped oscillations resonance

Two point masses $m _1$ and $m _2$ are coupled by a spring of spring. Constant $k$ and uncompressed length $L _0$. The spring is fully compressed and a thread ties the masses together with negligible separation between them. The tied assembly is moving in the $+x$ direction with uniform speed $v _0$. At a time, say $t = 0$, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring, start oscillating. The displacement of mass $m _1$ given by $x _1(t) = v _0 t(1 - cos \omega t)$ where $A$ is a constant. Find (i) the displacement $x _2(t)$ is $m _2$, and (ii) the relationship between $A$ and $L _0$.

  1. (i) $v _0 t + \dfrac{m _1}{2m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{2m _1 + m _2}\right)$
  2. (i) $v _0 t + \dfrac{m _1}{m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{m _1 + m _2}\right)$
  3. (i) $v _0 t + \dfrac{m _1}{3m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{3m _1 + m _2}\right)$
  4. (i) $v _0 t + \dfrac{m _1}{4m _2}A(1 - cos \omega t)$

    (ii) $A = \left(\dfrac{m _2}{4m _1 + m _2}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using conservation of momentum and the properties of a spring-mass system, the center of mass velocity remains constant. The displacement expressions are derived from the relative motion of the two masses oscillating about the center of mass.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

The natural angular frequency of a particle of mass 'm' attached to an ideal spring of force constant 'K' is

  1. $\sqrt{\frac{K}{m}}$
  2. $\sqrt{\frac{m}{K}}$
  3. $\left ( \frac{K}{m} \right )^{2}$
  4. $\left ( \frac{m}{K} \right )^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Suppose you displace the particle by a distance $'x'$
The spring now exerts a force,
This provides nccenary force for $SHM$
$\Rightarrow \ F=mwe^2x=k2$ ($w:$ natural angular frequency )
$\Rightarrow \ w=\sqrt {K/m}$