Tag: oscillation and waves

Questions Related to oscillation and waves

The net external force acting on the disc when its centre of mass is at displacement $x$ with respect to its equilibrium position is:

  1. $-kx$

  2. $-2kx$

  3. $-\dfrac{2kx}{3}$

  4. $\dfrac{4kx}{3}$


Correct Option: D

A particle is in S.H.M of amplitude $ 2$ cm. At extreme position the force is $4$N. At the point mid-way between mean and extreme position, the force is :

  1. $1$ N

  2. $2$N

  3. $3$N

  4. $4$N


Correct Option: B
Explanation:

Amplitude = 2 cm
Force = 4N
$F = mw^{2}A=4$
$F _{1} = mw^{2}x$

Since $F$ is directly proportional to $x$ so , at midpoint the force when the amplitude is $2 \ cm$ will be $2N$

A 1 kg mass executes SHM with an amplitude 10 cm, it takes $2\pi$ seconds to go from one end to the other end. The magnitude of the force acting on it at any end is :

  1. 0.1 N

  2. 0.2 N

  3. 0.5 N

  4. 0.05 N


Correct Option: A
Explanation:

As $  w  = \cfrac{2\pi}{T} = 1 \ rad/sec $ ;    Amplitude  $A  = 0.1 m$
magnitude of  force $ = m \times w^{2}.A$
                                  $=  0.1 N$

An elastic ball of density $d$ is released and it falls through a height $h$ before striking the surface of liquid of density $\rho(d < \rho)$. The motion of ball is:

  1. Periodic

  2. S.H.M.

  3. Circular

  4. Parabolic


Correct Option: A

A body of mass 1/4 kg is in S.H.M and its displacement is given by the relation $y= 0.05 sin(20t+\dfrac{\pi }{2})$ m. If $t$ is in seconds, the maximum force acting on the particle is:

  1. $5$ N

  2. $2.5$ N

  3. $10$ N

  4. $0.25$ N


Correct Option: A
Explanation:

$F= m\omega^{2}A$
$\omega = 20   rad / sec$
$A =   0.05   m$
Thus
$F= \dfrac{1}{4}\times 20\times 20\times \dfrac{1}{20}$
$=5 N $

If the energy density and velocity of a wave are $u$ and $c$ respectively then the energy propagating per second per unit area will be

  1. $u/c$

  2. $c^2u$

  3. $uc$

  4. $c/u$


Correct Option: C
Explanation:

If the energy density and velocity of a wave are $u$ and $c$ respectively then the energy propagating per second per unit area will be $uc.$

The kinetic energy per unit length for a wave on a string is the positional coordinate

  1. True

  2. False


Correct Option: B
Explanation:

The kinetic energy for a particle is given by $(\mu \Delta x/ 2) (\dfrac{dy}{dt})^2$

Thus, it depends only on the time variable and not on the position variable

A travelling wave has an equation of the form $A(x,t)=f(x+vt)$. The relation connecting positional derivative with time derivative of the function is:

  1. $\dfrac{dA}{dt}=\pm v^2 \dfrac {dA}{dx}$

  2. $\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$

  3. $\dfrac{dA}{dt}=\pm \sqrt(v) \dfrac {dA}{dx}$

  4. $\dfrac{dA}{dt}=(2 \pi v/\lambda) \dfrac {dA}{dx}$


Correct Option: B
Explanation:

Positional derivative and time derivative of a function f is $\dfrac{dA}{dt}=\pm v \dfrac {dA}{dx}$

The correct option is (b)

Kinetic energy per unit length for a particle in a standing wave is zero at:

  1. nodes

  2. antinodes

  3. mid-way between a node and an antinode

  4. None of the above


Correct Option: B
Explanation:

Particle at antinodes is momentarily at rest and hence has zero kinetic energy. Its speed comes down to zero at this point and all energy is stored in the form of potential energy.

The total energy per unit length for a travelling wave in a string of mass density $\mu$ , whose wave function is $A(x,t) = f(x \pm vt)$ is given by: 

  1. $E _tot = \sqrt(\mu/2) (\dfrac{dA}{dt})^2$

  2. $E _tot = (\mu/2) (\dfrac{dA}{dt})^2$

  3. $E _tot = (\mu/2)^2 (\dfrac{dA}{dt})^2$

  4. $E _tot = (2\mu) (\dfrac{dA}{dt})^2$


Correct Option: B