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,
$\omega =\frac{2\mathrm{\pi }}{\mathrm{T}}=\frac{2\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{3}{\sec }^{-1}$

Consider the equation of motion of S.H.M,
Y = Asin $\left(\omega t+\varphi \right)$ ...(1)
where Y is displacement of the particle, and
$\varphi$ 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

$$\begin{gathered} \sin \mathrm{\varphi }=\frac{1}{2} \\ \Rightarrow \mathrm{\varphi }=\frac{\mathrm{\pi }}{6} \end{gathered}$$

∴ Equation of displacement can be written as,
$x=\left(10\mathrm{cm}\right)\sin \left(\frac{\mathrm{\pi }}{3}t+\frac{\mathrm{\pi }}{6}\right)$

(ii) At t = 4 s,
$$\begin{gathered} x=10\sin \left[\frac{\mathrm{\pi }}{3}4+\frac{\mathrm{\pi }}{6}\right] \\ =10\sin \left[\frac{8\mathrm{\pi }+\mathrm{\pi }}{6}\right] \\ =10\sin \left(\frac{9\mathrm{\pi }}{6}\right) \\ =10\sin \left(\frac{3\mathrm{\pi }}{2}\right) \\ =10\sin \left(\pi +\frac{\mathrm{\pi }}{2}\right) \\ =-10\sin \frac{\mathrm{\pi }}{2}=-10 \end{gathered}$$

Acceleration is given by,
a = −ω²x
$$\begin{gathered} =\left(\frac{-{\mathrm{\pi }}^2}{9}\right)\times \left(-10\right) \\ =10.9\approx 11\mathrm{cm}/{\sec }^{-2} \end{gathered}$$