Question

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitude 2 cm, 1 m s⁻¹ and 10 m s⁻² at a certain instant. Find the amplitude and the time period of the motion.

Answer 1

It is given that:
Position of the particle, x = 2 cm = 0.02 m
Velocity of the particle, v = 1 ms^−1.
Acceleration of the particle, a = 10 ms^−2.

Let $\omega$ be the angular frequency of the particle.
The acceleration of the particle is given by,
a = ω²x
$$\begin{gathered} \Rightarrow \omega =\sqrt{\frac{a}{x}}=\sqrt{\frac{10}{0.02}} \\ =\sqrt{500}=10\sqrt{5}\mathrm{Hz} \\ \mathrm{Time}\mathrm{period}\mathrm{of}\mathrm{the}\mathrm{motion}\mathrm{is}\mathrm{given}\mathrm{as}, \\ T=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=\frac{2\mathrm{\pi }}{10\sqrt{5}} \\ =\frac{2\times 3.14}{10\times 2.236} \\ =0.28\mathrm{s} \end{gathered}$$

Now, the amplitude A is calculated as,
$$\begin{gathered} v=\omega \sqrt{A^2-x^2} \\ \Rightarrow v^2={\omega }^2\left(A^2-x^2\right) \\ 1=500\left(A^2-0.0004\right) \\ \Rightarrow A=0.0489=0.049\mathrm{m} \\ \Rightarrow A=4.9\mathrm{cm} \end{gathered}$$