Tag: sound: production of sound

Questions Related to sound: production of sound

"The motion of a particle with a restoring force gives oscillatory motion". Which of the following forces can be a restoring force for the oscillatory motion.

  1. Frictional force

  2. A constant force F

  3. A time varying force opposite to direction of motion

  4. Force due to a spring


Correct Option: D
Explanation:

Frictional force is a constant force acting opposite to the direction of motion
A constant force only makes an object to move in one direction
A time varying force makes an object to move in one direction with its speed increasing
Force due to a spring, keeps varying, as the object is moved away from its mean position and thus acts as a restoring force

The correct option is (d)

In periodic motion, the displacement is

  1. directly proportional to the restoring force

  2. inversely proportional to the restoring force

  3. independent of restoring force

  4. independent of any force acting on the particle


Correct Option: A
Explanation:

The displacement of the body from its equilibrium is directly proportional to the restoring force. Therefore, higher the displacement, higher is the restoring force.

The correct option is (a)

A thin spherical shell of mass $M$ and radius $R$ has a small hole. A particle of mass $m$ is released at its mouth. Then

  1. the particle will execute SHM inside the shell

  2. the particle will oscillate inside the shell, but the oscillations are not simple harmonic

  3. the particle will not oscillate, but the speed of the particle will go on increasing

  4. none of these


Correct Option: D
Explanation:

Firstly ,given that it is a  thin shell i.e hollow sphere then it will perform oscillation on SHM.

In case of solid sphere it performs SHM.

On the superposition of the two waves given as $y _1=A _0 \sin (\omega t-kx)$ and $y _2=A _0\cos \left(\omega t-kx+\dfrac{\pi}{6}\right) $the resultant amplitude of oscillations will be 

  1. $\sqrt{3}A _0$

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

  3. $A _0$

  4. $\dfrac{3}{2}A _0$


Correct Option: B

A particle oscillating in simple harmonic motion is :

  1. Never in equilibrium because it is in motion

  2. Never in equilibrium because there is always a force

  3. In equilibrium at the ends of its path because of zero velocity there

  4. In equilibrium at the centre of its path because the acceleration is zero there


Correct Option: A

A block of mass $M$ is performing $SHM$ with amplitude $A$ on a smooth horizontal surface$.$ At the extreme position a small block of mass $m$ falls vertically and sticks to$M.$ then$,$ amplitude of oscillation will be                                               

  1. $A$

  2. $A\sqrt {\dfrac{M}{{M + m}}} $

  3. $A\left( {\dfrac{M}{{M + m}}} \right)$

  4. $A\left( {\dfrac{{M + m}}{m}} \right)$


Correct Option: B

A simple harmonic oscillator of angular frequency $2$ rad/s 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 + \cfrac{1}{2} \cos 2t$

  2. $ \cos t - \cfrac{1}{2} \sin 2t$

  3. $ \sin t + \cfrac{1}{2} \sin 2t$

  4. $ \sin t - \cfrac{1}{2} \sin 2t$


Correct Option: D

Select proper wave equation which describes simple harmonic progressive wave travelling along positive $X$ axis.

  1. $y = A \sin ( \alpha t + k x )$

  2. $y = A \cos ( o t + k x )$

  3. $y = A \sin ( a t - k x )$

  4. $y = - 4 \tan ( \alpha x - k x )$


Correct Option: C

The motion represented by equation $x=2\sin \omega t+3\sin^{2}\omega t$ is

  1. Periodic

  2. Oscillatory

  3. $SHM$

  4. Both (1) & (2)


Correct Option: B

A particle of mass $0.1 kg$ executes SHM under a force $F = (-10 x) N$. Speed of particle at mean position is $6 m/s$. Then amplitude of oscillations is 

  1. $0.6 m$

  2. $0.2 m$

  3. $0.4 m$

  4. $0.1 m$


Correct Option: A
Explanation:

$\begin{array}{l} M=0.1\, Kg \ f=-10x \ \Rightarrow k=10 \ w=\sqrt { \frac { k }{ m }  }  \ =\sqrt { \frac { { 10 } }{ { 0.1 } }  } =10 \ { V _{ \max   } }=6m/s \ \Rightarrow A=\frac { 6 }{ { 10 } } =0.6m \ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$