Tag: force in shm

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Multiple choice force in shm oscillations oscillation and waves physics

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $2N$

Multiple choice force in shm oscillations oscillation and waves physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice force in shm oscillations oscillation and waves physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the ball hits the liquid, it experiences a buoyant force greater than its weight (since d < rho), causing it to decelerate and eventually rise. It will oscillate between the surface and the depth, making the motion periodic, but it is not SHM because the forces are not linear with displacement.

Multiple choice force in shm oscillations oscillation and waves physics

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $