Oscillations - Result Question 36
40. A particle executes linear simple harmonic motion with an amplitude of $3 cm$. When the particle is at $2 cm$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
[2017]
(a) $\frac{\sqrt{5}}{2 \pi}$
(b) $\frac{4 \pi}{\sqrt{5}}$
(c) $\frac{2 \pi}{\sqrt{3}}$
(d) $\frac{\sqrt{5}}{\pi}$
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Answer:
Correct Answer: 40. (b)
Solution:
- (b) Given, Amplitude $A=3 cm$
When particle is at $x=2 cm$
According to question, magnitude of velocity $=$ acceleration
$\omega \sqrt{A^{2}-x^{2}}=x \omega^{2}$
$\sqrt{(3)^{2}-(2)^{2}}=2(\frac{2 \pi}{T})$
$\sqrt{5}=\frac{4 \pi}{T} \Rightarrow T=\frac{4 \pi}{\sqrt{5}}$
