Courses for Kids
Free study material
Offline Centres
Store Icon

A particle executes S.H.M. with an amplitude of 10 cm and period of 10 s. Find the (i) velocity (ii) acceleration of the particle at a distance 5 cm from the equilibrium position.

Last updated date: 05th Mar 2024
Total views: 362.1k
Views today: 9.62k
IVSAT 2024
362.1k+ views
Hint: In case of simple harmonic motion there will be a mean position and two extreme positions. The pendulum executes the SHM with respect to the mean position and within the range of two extreme positions. Generally simple harmonic motions are denoted with the sinusoidal or cosecant functions.
Formula used:
  & x = A\sin (\varphi ) \cr
  & \varphi = \omega t \cr} $

Complete answer:
In SHM acceleration of the pendulum executing the SHM is proportional to the displacement of the pendulum from the mean position. Generally if an object starts from the mean position then we write displacement as a sinusoidal function and if an object starts from the extreme position then we write the displacement of a body as a cosine function. We can assume in any way if it is not specified in the question. It's given in the question that pendulum is executing the SHM about the mean position. Then the function of the SHM displacement will be $x = A\sin (\varphi )$ where $\varphi $ is the phase of the pendulum.
Amplitude will be maximum at extreme position and minimum at the mean position. The phase difference between the successive extreme and mean positions will be 90 degrees.
Let us assume SHM equation for particle is
  & x = A\sin (\varphi ) \cr
  & \Rightarrow \varphi = \omega t \cr} $
$\therefore \omega = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{{10}} = 0.6283$
Where T is the time period and A is the amplitude
Now acceleration(a) is given by
$a = - {\omega ^2}x$
Where x is the distance from the mean position
$a = - {\omega ^2}x$
  & \Rightarrow a = - {\left( {0.6283} \right)^2}5 \cr
  & \therefore a = - 1.974m/{s^2} \cr} $
Velocity is given as
$v = \omega \sqrt {{A^2} - {x^2}} $
  & \Rightarrow v = 0.6283\sqrt {{{10}^2} - {5^2}} \cr
  & \therefore v = \pm 5.442cm/s \cr} $
Hence we had got the values of acceleration and velocity at 5cm from the mean position.

In the acceleration we have negative sign because if the particle moves away from the mean position, acceleration acts towards the mean position and if the particle moves towards the mean position, acceleration acts away from the mean position. So it is always opposite to displacement in SHM.
Recently Updated Pages
Trending doubts