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) $\frac{\sqrt{5}}{\pi}$

Show Answer

Answer:

Correct Answer: 40. (b)

Solution:

  1. (b) Given, Amplitude $A=3 cm$

When particle is at $x=2 cm$

According to question, magnitude of velocity $=$ acceleration

$\omega \sqrt{A^{2}-x^{2}}=x \omega^{2}$

$\sqrt{(3)^{2}-(2)^{2}}=2(\frac{2 \pi}{T})$

$\sqrt{5}=\frac{4 \pi}{T} \Rightarrow T=\frac{4 \pi}{\sqrt{5}}$



NCERT Chapter Video Solution

Dual Pane