A particle executes simple harmonic oscillation with an amplitude… - Vidyadip

MCQ

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
$\frac{T}{8}$
✓
$\;\frac{T}{{12}}$
C
$\;\frac{T}{2}$
D
$\;\frac{T}{4}$

✓

Answer

Correct option: B.

$\;\frac{T}{{12}}$

b
$x(t)=a \sin \omega t$ $(from \,the\, equilibrium\, position)$

At $x(t)=a / 2$

$\therefore \quad \frac{a}{2}=a \sin (\omega t)$

$\Rightarrow \sin \left(\frac{\pi}{6}\right)=\sin (\omega t) \quad$ or, $\quad \frac{\pi}{6}=\frac{2 \pi t}{T}\left[\because \omega=\frac{2 \pi}{T}\right]$

or $t=T / 12$

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