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

[EAMCET (Med.) 2000]

• A. T
• B. T/4
• C. T/8
• D. T/16

Answer 1

The correct option is C

T/8

y = a sin $\frac{2\pi }{T}t\Rightarrow \frac{a}{\sqrt{2}}asin\frac{2\pi }{T}.t$

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