Simple Harmonic Motion 3 Question 3
3. A particle executes simple harmonic motion with an amplitude of $5 cm$. When the particle is at $4 cm$ from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is
(2019 Main, 10 Jan II)
(a) $\frac{4 \pi}{3}$
(b) $\frac{8 \pi}{3}$
(c) $\frac{7}{3} \pi$
(d) $\frac{3}{8} \pi$
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Answer:
Correct Answer: 3. (b)
Solution:
- In simple harmonic motion, position $(x)$, velocity $(v)$ and acceleration $(a)$ of the particle are given by
$$ \begin{aligned} & x=A \sin \omega t \\ & v=\omega \sqrt{A^{2}-x^{2}} \text { or } v=A \omega \cos \omega t \end{aligned} $$
$$ \text { and } \quad a=-\omega^{2} x \quad \text { or } a=-\omega^{2} A \sin \omega t $$
Given, amplitude $A=5 cm$ and displacement $x=4 cm$. At this time (when $x=4 cm$ ), velocity and acceleration have same magnitude.
$$ \begin{aligned} & \Rightarrow \quad\left|v _{x=4}\right|=\left|a _{x=4}\right| \text { or }\left|\omega \sqrt{5^{2}-4^{2}}\right|=\left|-4 \omega^{2}\right| \\ & \Rightarrow \quad 3 \omega=+4 \omega^{2} \Rightarrow \omega=(3 / 4) rad / s \end{aligned} $$
So, time period, $T=\frac{2 \pi}{\omega} \Rightarrow T=\frac{2 \pi}{3} \times 4=\frac{8 \pi}{3} s$
