Tag: free, forced and damped oscillations

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Questions Related to free, forced and damped oscillations

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

In forced oscillation displacement equation is $x(t)=A\cos(\omega _{d}t+\theta)$ then amplitude $'A'$ vary with forced angular frequency $\omega _{d}$ and natural angular frequency $'\omega'$ as (b=dumping constant)

  1. $\dfrac{F}{m\omega^{2}}$
  2. $\dfrac{F}{\left\{m^{2}(\omega^{2}-\omega _{d}^{2})^{2}+\omega _{d}^{2}b^{2}\right\}^{1/2}}$
  3. $\dfrac{F}{m(\omega^{2}-\omega _{d}^{2})}$
  4. $\dfrac { F }{ { \left\{ m\left( { \omega } _{ d }^{ 2 }{ b }^{ 2 } \right) +\left( { \omega }^{ 2 }-{ \omega } _{ d }^{ 2 } \right) \right\} }^{ 1/2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In forced oscillations, the amplitude A of a damped harmonic oscillator driven by a periodic force F is given by A = F / sqrt(m^2(omega^2 - omega_d^2)^2 + omega_d^2 b^2), where omega is the natural frequency, omega_d is the driving frequency, and b is the damping constant.

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

In damped oscillation, the amplitude of oscillation is reduced to 1/3 of its initial value $A _0$ at the end of 100 oscillations. When the system completes 200 oscillations, its amplitude must be

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

Amplitude follows A(n) = A0 * r^n. After 100 oscillations, A(100) = A0/3. After 200 oscillations, A(200) = A0 * (r^100)^2 = A0 * (1/3)^2 = A0/9.

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

If ${ \omega  } _{ 0 }$ is natural frequency of damped forced oscillation and p that of driving force, then for amplitude resonance

  1. ${ p } _{ r }={ \omega } _{ 0 }$
  2. ${ p } _{ r }<{ \omega } _{ 0 }$
  3. ${ p } _{ r }>{ \omega } _{ 0 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a damped forced oscillator, the amplitude resonance occurs at a driving frequency p_r = sqrt(omega_0^2 - b^2/2m^2), which is less than the natural frequency omega_0.

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

A pendulum with time of 1 s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in $s^{-1}$) is

  1. 2
  2. $\dfrac{1}{15} ln 3$
  3. $\dfrac{1}{2}$
  4. $\dfrac{1}{30} ln 3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy E(t) = E0 * exp(-2kt). E(15) = 15, E0 = 45. 15 = 45 * exp(-2k * 15). 1/3 = exp(-30k). ln(1/3) = -30k. k = ln(3)/30.

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

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

$A = {A _0}{e^{ - kt}}$

$0.9{A _0} = {A _0}{e^{ - kt}}$
$ - kt = \ln \left( {0.9} \right) \Rightarrow  - 15k = 3\ln \left( {0.9} \right)$
$A = {A _0}{e^{ - 15k}} = {A _0}{e^{ - ln{{\left( {0.9} \right)}^3}}}$
$ = {\left( {0.9} \right)^3}{A _0} = 0.729{A _0}$
Hence,
option $(C)$ is correct answer.

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

A mass of 50 kg is suspended from a spring of stiffness 10 kN/m. It is set oscillating and it is observed that two successive oscillations have amplitudes of 10 mm and 1 mm. Determine the damping ratio.

  1. 0.315

  2. 0.328

  3. 0.344

  4. 0.353
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation


For successive amplitudes $m = 1$
amplitude reduction factor

$=ln\left( \dfrac { { x } _{ 1 } }{ { x } _{ 2 } }  \right) =ln\left(

\dfrac { 10 }{ 1 }  \right) =ln10=2.3$
amplitude reduction factor $=\dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  } $
$\Rightarrow \dfrac { 2\pi \delta m }{ \sqrt { 1-{ \delta  }^{ 2 } }  } =2.3$
squaring both sides
$\dfrac { 39.478{ \delta  }^{ 2 } }{ 1-{ \delta  }^{ 2 } } =5.29\\ \Rightarrow { \delta  }^{ 2 }=0.118\\ \Rightarrow \delta =0.344$

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.