Tag: free, forced and damped oscillations

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Questions Related to free, forced and damped oscillations

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

Assertion (A): In damped vibrations, amplitude of oscillation decreases
Reason (R): Damped vibrations indicate loss of energy due to air resistance

  1. Both A and R are true and R is the correct explanation of A

  2. Both A and R are true and R is not the correct explanation of A

  3. A is true and R is false

  4. A is false and R is true
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Damped vibrations in which an oscillating system has the effect of reducing, restricting or preventing its oscillations.

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

A particle with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force $F \ sin \omega.$ If the amplitude of the particle is maximum for $\omega = \omega _1$ and the energy of the particle is maximum for $\omega = \omega _2$ then (where $\omega _0$ natural frequency of oscillation of particle)

  1. $\omega _1 = \omega _0 \ and \ \omega _2 \neq \omega _0$
  2. $\omega _1 = \omega _0 \ and \ \omega _2 = \omega _0$
  3. $\omega _1 \neq \omega _0 \ and \ \omega _2 =\omega _0$
  4. $\omega _1 \neq \omega _0 \ and \ \omega _2 \neq \omega _0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As we know the energy of the particle is maximum at natural frequency. Since, the restoring force is proportional to displacement and resisting force is proportional to velocity. So the correct option is ${{\omega } _{0}}={{\omega } _{2}}\,\And \,{{\omega } _{1}}\ne\,{{\omega } _{0}}$

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

Few particles undergo damped harmonic motion. Values for the spring constant $k$ , the damping constant $b$ , and the mass $m$ are given below. Which leads to the smallest rate of loss of mechanical energy at the initial moment?

  1. $ k = 100N/m , m = 50 g, b = 8 g/s $
  2. $ k = 150 N/m , m = 50 g, b = 5 g/s $
  3. $ k = 150N/m , m = 10g, b = 8 g/s $
  4. $ k = 200N/m , m = 8g, b = 6 g/s $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The rate of loss of mechanical energy is proportional to the damping force times velocity, or P = b * v^2. For a given initial displacement, the initial velocity is zero, but the damping force acts as the system moves. The damping coefficient b is the primary factor. Comparing the options, the smallest b value (5 g/s) leads to the smallest rate of energy loss.

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

A bar magnet oscillates with a frequency of$ 10 $ oscillations per minute. When another bar magnet is placed on its axis at a small distance, it oscillates at $14$ oscillations per minute. Now, the second bar magnet is turned so that poles are instantaneous, keeping the location same. The new frequency of oscillation will be 

  1. $2$ vibrations/min
  2. $4$ vibrations/min
  3. $10$ vibrations/min
  4. $14$ vibrations/min
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{60}{10}= 2\pi \sqrt{\dfrac{l}{MB _H}}$
$\dfrac{60}{14}= 2\pi \sqrt{\dfrac{l}{M(MB _H)}}$
$\therefore \dfrac{7}{5} = \sqrt{B _H +B}{B _H} $ or $ B= \dfrac{24}{25}B _H$
Hence,
$\dfrac{60}{10}= 2\pi \sqrt{\dfrac{l}{M(B _H-B)}}= 2\pi \sqrt{\dfrac{l}{MB(1-24/25)}}$
$= 5\times 2\pi \sqrt{\dfrac{l}{2MB}} = 5 \times \dfrac{60}{10}= 30$
$\therefore f= \dfrac{60}{30} = 2$ vibrations/ min

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

The angular frequency of the damped oscillator is given by $\omega =\sqrt{\left(\frac{k}{m} -\dfrac{r^2}{4m^2}\right)}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\dfrac{r^2}{mk}$ is $8%$, the changed in time period compared to the undamped oscillator is approximately as follows:  

  1. Increases by 1%

  2. Decreases by 1%

  3. Decreases by 8%

  4. increases by 8%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\omega = \sqrt {\dfrac{k}{m}-\dfrac{r^2}{4m^2}}= \sqrt{\dfrac{k}{m}}\sqrt{1-\dfrac{r^2}{4mk}}$
 $\approx \omega _o \left(1-\dfrac{r^2}{8mk}\right) \approx$ (1 - 1%)

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

The amplitude of a damped oscillator becomes $\left (\dfrac {1}{3}\right )rd$ in $2s$. If its amplitude after $6\ s$ in $\dfrac {1}{n}$ times the original amplitude, the value of $n$ is

  1. $3^{2}$
  2. $3\sqrt {2}$
  3. $3^{3}$
  4. $2^{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let original amplitude $=A$

Amplitude after 2 sec=$\dfrac{A}{3}$
Amplitude after next 2 sec=$\dfrac{A}{3}\times \dfrac{1}{3}=\dfrac{A}{9}$
Amplitude again  after 2 sec=$\dfrac{1}{3}\times \dfrac{A}{9}=\dfrac{A}{27}=\dfrac{A}{3^3}$
Here $n=3^3$

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.