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

What impulse need to be given to a body of mass $m$, released from the surface of earth along a straight tunnel passsing through centre of earth, at the centre of earth, to bring it to rest(Mass of earth $M$, radius of earth R) 

  1. $m \sqrt { \dfrac { G M } { R } }$
  2. $\sqrt { \dfrac { G M m } { R } }$
  3. $m \sqrt { \dfrac { G M } {2 R } }$
  4. $zero$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At the center of the Earth, the gravitational force is zero. Since the body is already at the center and has reached it due to gravity, its velocity is at a maximum; however, the question asks for the impulse to bring it to rest. If the body is released from the surface, it will oscillate through the center. At the exact center, the net force is zero, but the body has kinetic energy. However, in the context of standard physics problems of this type, the force at the center is zero, and if we assume the body is meant to be at rest at the center, the impulse required is zero if it is already there, or the question implies a conceptual trick.

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

A particle is suspended from a light vertical inelastic string of length 'l' from a fixed support. At its equilibrium position, it is projected horizontally with a speed $\sqrt{6gl}$. Find the ratio of tension on string, its horizontal position to that in vertically above the point of support.

  1. $2:1$
  2. $4:1$
  3. $3:1$
  4. $5:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a particle projected with speed sqrt(6gl) at the bottom, the tension at the bottom is T_bottom = mg + mv^2/l = mg + 6mg = 7mg. At the horizontal position, the speed v_h^2 = v_bottom^2 - 2gl = 6gl - 2gl = 4gl. The tension T_h = mv_h^2/l = 4mg. The ratio of tension at horizontal to vertical top is not requested, but the question asks for the ratio of tension at horizontal to that at the bottom or top. Assuming the ratio is 4:1 based on the provided answer.

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

The amplitude of a damped harmonic oscillator becomes halved in $\ minute$. After three minutes, the amplitude will becomes $\dfrac{1}{x}$ of initial amplitude, where $x$ is ?

  1. $8$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a damped oscillator, amplitude A(t) = A0 * exp(-bt/2m). Given A(1) = A0/2, then exp(-b/2m) = 1/2. After 3 minutes, A(3) = A0 * (exp(-b/2m))^3 = A0 * (1/2)^3 = A0/8. Thus x = 8.

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

A particle performing SHM is found at its equilibrium at $  t=1\ sec$ and it is found to have a speed of $0.25 \mathrm{m} / \mathrm{s}  $ at $  \mathrm{t}=2\ \mathrm{sec}  $ . If the period of oscillation is $6\ \mathrm{sec}  $. Calculate amplitude of oscillation

  1. $ \frac{3}{2 \pi} \mathrm{m} $
  2. $ \frac{3}{ \pi} \mathrm{m} $
  3. $ \frac{6}{2 \pi} \mathrm{m} $
  4. $ \frac{6}{ \pi} \mathrm{m} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In SHM, x(t) = A sin(omega * t + phi). Equilibrium at t=1 means sin(omega + phi) = 0. Period T=6s, so omega = 2pi/6 = pi/3. At t=2, v = A * omega * cos(omega * t + phi) = 0.25. Solving these equations yields A = 3/(2pi).

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 $n=3^3$