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Q. The acceleration of a particle performing SHM is $12\, cm / s ^{2}$ at a distance of $3\, cm$ from the mean position. Its time period is

ManipalManipal 2011Oscillations

Solution:

Acceleration $a=\omega^{2} y$
$a =\left(\frac{2\, \pi}{T}\right)^{2} y$
$\Rightarrow T^{2} =\frac{4 \pi^{2} y}{a}$
$\Rightarrow T=2\, \pi \sqrt{\frac{y}{a}}$
$=2\, \pi \sqrt{\frac{3}{12}}$
$T =2\, \pi \times \frac{1}{2}=3.14\, s$