Question

A particle executes simple harmonic motion with an amplitude of $ 10cm $ and time period $ 6s $ . At $ t = 0 $ it is at position $ x = 5cm $ from mean position and going towards positive x-direction. Write the equation for the displacement $ x $ at time $ t $ . Find the magnitude of the acceleration of the particle at $ t = 4s $ .

Answer 1

Hint :In order to obtain the equation of displacement of $ x $ with respect to the time $ t $ , we have to first find the angular frequency of the simple harmonic motion. Then by using this equation of displacement, we can calculate the magnitude of acceleration of the particle at $ t = 4s $ .
 $
  \omega = \dfrac{{2\pi }}{T} \\
  Y = A\sin (\omega t + \phi ) \\
  a = - {\omega ^2}x \\
  $

Complete Step By Step Answer:
From the question, we know that the amplitude of simple harmonic motion, $ A = 10cm $ and the time period of the simple harmonic motion, $ T = 6s $ .
When the time is at $ t = 0 $ , the particle is at a position $ x = 5cm $ from the mean position.
In order to obtain the equation of displacement of $ x $ with respect to the time $ t $ , we have to first find the angular frequency of the simple harmonic motion.
Angular frequency $ (\omega ) $ is given by,
 $ \omega = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{6} = \dfrac{\pi }{3}{\sec ^{ - 1}} $
We have obtained the angular frequency which is $ \omega = \dfrac{\pi }{3}{\sec ^{ - 1}} $
Now, let’s consider the equation of simple harmonic motion which is given by,
 $ Y = A\sin (\omega t + \phi ) $ [let this be equation (1)]
where $ Y $ is the displacement of the particle, and $ \phi $ is the phase of the particle.
Now, we shall substitute the values of $ A,t $ and $ \omega $ in equation (1).
Then we get,
 $
  5 = 10\sin (\omega \times 0 + \phi ) \\
   \Rightarrow 5 = 10\sin \phi \\
   \Rightarrow \sin \phi = \dfrac{1}{2} \\
   \Rightarrow \phi = \dfrac{\pi }{6} \\
  $
Therefore, the equation of displacement can be written as,

 $ x = (10cm)\sin (\dfrac{\pi }{3}t + \dfrac{\pi }{6}) $
Now, we have to find the magnitude of the acceleration of the particle at $ t = 4s $ . So, we substitute the value $ t = 4 $ in the equation of displacement.
 $ x = (10)\sin (\dfrac{\pi }{3}(4) + \dfrac{\pi }{6}) $
By taking LCM, we get,
 $
  x = 10\sin (\dfrac{{8\pi + \pi }}{6}) \\
   = 10\sin (\dfrac{{9\pi }}{6}) \\
   = 10\sin (\dfrac{{3\pi }}{2}) \\
   = 10\sin (\pi + \dfrac{\pi }{2}) \\
  $
 Since $ \sin (\pi + \phi ) = - \sin (\phi ) $
We get,
 $ x = - 10\sin (\dfrac{\pi }{2}) $
Since $ \sin (\dfrac{\pi }{2}) = 1 $
 $ x = - 10 $
Therefore, we found that the displacement of the particle, $ x = - 10cm $
The magnitude of acceleration of the particle is given by the formula,
 $ a = - {\omega ^2}x $
Now, we substitute the values of $ \omega $ and $ x $ in the above equation and we get,
 $
  a = - {(\dfrac{\pi }{3})^2} \times ( - 10) \\
   = (\dfrac{{{\pi ^2}}}{9}) \times 10 \\
   = 10.95 \\
   \approx 11 \\
  $
 Therefore, the magnitude of acceleration of the particle is $ a \approx 11cm.{\sec ^{ - 2}} $ .

Note :
If the magnitude of acceleration obtained is a negative value, this indicates that the acceleration and the displacement are in the opposite direction of each other. If the magnitude of acceleration obtained is a positive value, this indicates that the acceleration and the displacement are in the same direction. So, the students have to be careful while solving, as a negative sign makes a lot of difference.