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

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be (intial amplitude =$A _{0}$)

  1. $\dfrac{1}{4}A _{0}$
  2. $\dfrac{1}{2}A _{0}$
  3. $\dfrac{1}{5}A _{0}$
  4. $\dfrac{1}{7}A _{0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In damped oscillations, damping force is directly proportional to speed to oscilator . If amplitude becomes half of its maximum value in 1s , then after 2 s amplitude will be

 

Amplitude is given by:

$A={{A} _{o}}{{e}^{-\alpha t}}$

Where, $A$ is amplitude at time t.

t is time

${{A} _{0}}$ is initial aplitude

$\alpha $ is constant

At t = 1s

$A=\dfrac{{{A} _{0}}}{2}$

So,

$ \dfrac{{{A} _{0}}}{2}={{A} _{0}}{{e}^{-\alpha }} $

$ {{e}^{-\alpha }}=\dfrac{1}{2} $

At t = 2s

$ A={{A} _{0}}{{e}^{-2\alpha }} $

$ A={{A} _{0}}{{(\dfrac{1}{2})}^{2}} $

$ A=\dfrac{{{A} _{0}}}{4} $

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

  1. $0.67\ J$
  2. $0.067\ J$
  3. $6.7\ J$
  4. $0.5\ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $0.067\,J$
Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

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 $2\ minutes$?

  1. $A _{0}/6$
  2. $A _{0}/9$
  3. $A _{0}/4$
  4. $A _{0}/3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A(t) = A0 * exp(-kt). Given A(6) = A0/27, so exp(-6k) = 1/27 = (1/3)^3. Thus exp(-2k) = 1/3. At t=2, A(2) = A0 * exp(-2k) = A0/3.

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

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?

  1. $0.81$
  2. $0.73$
  3. $0.95$
  4. $0.66$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Amplitude A(t) = A0 * exp(-kt). A(5) = 0.9 * A0, so exp(-5k) = 0.9. Energy E is proportional to A^2. E(10) = E0 * (A(10)/A0)^2 = E0 * (exp(-10k))^2 = E0 * (exp(-5k))^4 = E0 * (0.9)^4 = 0.6561 * E0. The closest option is 0.73, suggesting a potential calculation difference or rounding.

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