Question

A particle executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude.

• A. $\frac{T}{2}$
• B. $\frac{T}{4}$
• C. $\frac{T}{8}$
• D. $\frac{T}{12}$

Answer 1

The correct option is D $\frac{T}{12}$

The displacement equation $x=A$ $sinwt$ where $A$ is the amplitude of oscillation

The particle is at mean position initially i.e at $t=0$, $x=0$

At the instant $t$, the particle is at a position of half amplitude i.e $x=\frac{A}{2}$

$\therefore$ $\frac{A}{2}=A$ $sinwt$

$0.5=sinwt$ $⟹wt=\frac{\pi }{6}$

$\frac{2\pi }{T}\times t=\frac{\pi }{6}$ $⟹t=\frac{T}{12}$