Question

A particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

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

Answer 1

Answer: (C)

$\frac{T}{12}$

Answer 2

In simple harmonic motion, the displacement $x(t)$ of a particle from equilibrium position at any time $t$ is given by $x(t)=a\,sin\,\omega\,t$ where a is the amplitude At $x(t)=\frac{a}{2}$ $\frac{a}{2}=a\,sin\,\omega\,t$ or $\frac{1}{2}=sin\,\omega\,t$ or $sin\,30^{\circ}=sin\omega \,t$ or $sin\left(\frac{\pi}{6}\right)=sin\,\omega t$ $\frac{\pi}{6}=\frac{2\pi}{T} t (\because \omega=\frac{2\pi}{T}$, where T is the time perood of oscillation) or $t=\frac{T}{12}$