Tag: resonance

Questions Related to resonance

In damped oscillation mass is $1\ kg$ and spring constant $=100\ N/m$, damping coefficeint$=0.5\ kg\ s^{-1}$. If the mass displaced by $10\ cm$ from its mean position then what will be the value of its mechanical energy after $4$ seconds?

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

  1. $0.67\ J$

  2. $0.067\ J$

  3. $6.7\ J$

  4. $0.5\ J$

Show answer
Correct Option: B
Explanation:
Given,

Mass, $m=1\,kg$

Spring constant, $k=100\,N/m^2$

Damping coefficient, $b=0.5\,kg/s$

Distance, $x=10\,cm$

Time, $t=4\,s$

We know,

The energy for damped oscillation, $E=\dfrac 12kx^2 e^{-\dfrac{bt}{m}}$

$E=\dfrac 12\times 100\times 0.01\times e^{-\dfrac{0.5\times 4}{1}}$

$E=\dfrac{e^{-2}}{2}=0.067\,J$

Hence the mechanical energy is $0.067\,J$

The amplitude of a damped harmonic oscillator becomes $\left (\dfrac {1}{27}\right )^{th}$ of its initial value $A _{0}$ after $6$ minute. What was the amplitude after $2\ minutes$?

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

  1. $A _{0}/6$

  2. $A _{0}/9$

  3. $A _{0}/4$

  4. $A _{0}/3$

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Correct Option: D

The amplitude of a damped oscillator decreases to $0.9$ times its initial value in $5$ seconds. By how many times to its initial value, energy of oscillation decreases to, in $10$ seconds?

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

  1. $0.81$

  2. $0.73$

  3. $0.95$

  4. $0.66$

Show answer
Correct Option: B

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)

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

  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 } } $

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Correct Option: C

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

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

  1. $\dfrac{A _0}{2}$

  2. $\dfrac{A _0}{4}$

  3. $\dfrac{A _0}{6}$

  4. $\dfrac{A _0}{9}$

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Correct Option: D

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

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

  1. ${ p } _{ r }={ \omega } _{ 0 }$

  2. ${ p } _{ r }<{ \omega } _{ 0 }$

  3. ${ p } _{ r }>{ \omega } _{ 0 }$

  4. None of these

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Correct Option: B

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

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

  1. 2

  2. $\dfrac{1}{15} ln 3$

  3. $\dfrac{1}{2}$

  4. $\dfrac{1}{30} ln 3$

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Correct Option: B

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

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
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.

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.

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

  1. 0.315

  2. 0.328

  3. 0.344

  4. 0.353

Show answer
Correct Option: C
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
$\dfrac { 39.478{ \delta  }^{ 2 } }{ 1-{ \delta  }^{ 2 } } =5.29\\ \Rightarrow { \delta  }^{ 2 }=0.118\\ \Rightarrow \delta =0.344$

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