Question

A particle is executing simple harmonic motion with a period of T seconds and amplitude a metre. The shortest time it takes to reach a point $\frac{a}{\sqrt{2}}m$ from its mean position in seconds is

• A. T
• B. T/4
• C. T/8
• D. T/16

Answer 1

Answer: (C)

T/8

Answer 2

$y = a\sin\frac{2\pi}{T}t$ ⇒ $\frac{a}{\sqrt{2}} = a\sin\frac{2\pi}{T} \cdot t$

⇒ $\sin\frac{2\pi}{T}t = \frac{1}{\sqrt{2}} = \sin\frac{\pi}{4}$⇒ $\frac{2\pi}{T}t = \frac{\pi}{4}$⇒ $t = \frac{T}{8}$