Question

A particle executes simple harmonic motion with an amplitude of 10 cm and time period 6 s. At t = 0 it is at position x = 5 cm 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 = 4 s.

Answer 1

It is given,
Amplitude of the simple harmonic motion, A =10 cm
At t = 0 and  x = 5 cm,
Time period of the simple harmonic motion, T  = 6 s
Angular frequency (ω) is given by,

\[\omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} \sec^{- 1}\]

Consider the equation of motion of S.H.M,
  Y = Asin \[\left( \omega t + \phi \right)\]...(1)

where Y is displacement of the particle, and \[\phi\] is phase of the particle.

On substituting the values of A, t and ω in equation (1), we get:
 5 = 10sin(ω × 0 + ϕ)

\[\Rightarrow\] 5 = 10sin ϕ

\[\sin \phi = \frac{1}{2}\]

\[ \Rightarrow \phi = \frac{\pi}{6}\]

∴ Equation of displacement can be written as,

\[x = \left( 10 \text { cm }\right) \sin \left( \frac{\pi}{3}t + \frac{\pi}{6} \right)\]

(ii) At t = 4 s,

\[x = 10\sin\left[ \frac{\pi}{3}4 + \frac{\pi}{6} \right]\]

\[ = 10\sin\left[ \frac{8\pi + \pi}{6} \right]\]

\[ = 10\sin\left( \frac{9\pi}{6} \right)\]

\[ = 10\sin\left( \frac{3\pi}{2} \right)\]

\[ = 10\sin\left( \pi + \frac{\pi}{2} \right)\]

\[ = - 10\sin\frac{\pi}{2} = - 10\]

Acceleration is given by,
a = −ω²x

\[= \left( \frac{- \pi^2}{9} \right) \times \left( - 10 \right)\]

\[ = 10 . 9 \approx 11 \text { cm }/ \sec^{- 2}\]