Question

An instantaneous displacement of a simple harmonic oscillator is $x=A\cos (\omega t+\frac{\pi }{4})$. Its speed will be maximum at

• A. $\frac{\pi }{4\omega }$
• B. $\frac{\pi }{2\omega }$
• C. $\frac{\pi }{\omega }$
• D. $\frac{2\pi }{\omega }$

Answer 1

The correct option is A $\frac{\pi }{4\omega }$

Velocity, $v=\frac{dx}{dt}=-A\omega \sin (\omega t+\pi /4)$
Velocity will be maximum, when
$\omega t+\frac{\pi }{4}=\frac{\pi }{2}$
$\Rightarrow \omega t=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$
$\Rightarrow t=\frac{\pi }{4\omega }$