Question

A particle undergoes simple harmonic motion having time period $T.$Find the time taken in $\frac{3}{8}th$ oscillation.

Answer 1

Since, time period is T

$\Rightarrow$ angular frequency$\left(\omega \right)=\frac{2\pi }{T}$

We know that, simple harmonic motion(SHM) is the projected uniform circular motion of a particle taken on diameter of circle.

In circular motion,

$\theta =\omega t$

So, $1$ oscillation $\equiv 2\pi$ radian

${\left(\frac{3}{8}\right)}^{th}$ oscillation $\equiv \frac{3}{8}\left(2\pi \right)=\frac{3\pi }{4}$ rad

$\frac{3\pi }{4}=\omega t$

$\Rightarrow \frac{2\pi }{T}t=\frac{3\pi }{4}$

$\Rightarrow t=\frac{3T}{8}$.