Tag: force in shm

Questions Related to force in shm

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 $