First slide

Simple harmonic motion

Question

A particle executes SHM of amplitude A and time period T. The distance travelled by the particle in the duration its phase changes from $\frac{π}{12} to \frac{5 π}{12}$

Moderate

A

$\frac{1}{\sqrt{2}} A$
B

$\sqrt{\frac{3}{2}} A$
C

$\frac{2}{\sqrt{3}} A$
D

$\sqrt{\frac{2}{3}} A$

Solution

Distance travelled by the particle
$= A \sin \frac{5 π}{12} - A \sin \frac{π}{12} = \frac{A}{\sqrt{2}}$

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