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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.
Solution
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,
\[v = \omega\sqrt{A^2 - x^2}\]
\[ \Rightarrow v^2 = \omega^2 \left( A^2 - x^2 \right)\]
\[ 1 = 500\left( A^2 - 0 . 0004 \right)\]
\[ \Rightarrow A = 0 . 0489 = 0 . 049 m\]
\[ \Rightarrow A = 4 . 9 \text { cm }\]
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