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

\[\Rightarrow \omega = \sqrt{\frac{a}{x}} = \sqrt{\frac{10}{0 . 02}}\]

\[ = \sqrt{500} = 10\sqrt{5} Hz\]

\[\text { Time period of the motion is given as, } \]

\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{10\sqrt{5}}\]

\[ = \frac{2 \times 3 . 14}{10 \times 2 . 236}\]

\[ = 0 . 28 s\]

Now, the amplitude A is calculated as,

\[v = \omega\sqrt{A^2 - x^2}\]

\[ \Rightarrow v^2 = \omega^2 \left( A^2 - x^2 \right)\]

\[ 1 = 500\left( A^2 - 0 . 0004 \right)\]

\[ \Rightarrow A = 0 . 0489 = 0 . 049 m\]

\[ \Rightarrow A = 4 . 9 \text { cm }\]