First slide

Simple harmonic motion

Question

A particle executes simple harmonic motion (amplitude = A) between $x = - A$ and $x = + A$ . The time taken for it to go from 0 to A/2 is $T_{1}$ and to go from A/2 to A is $T_{2}$ . Then

Moderate

A

$T_{1} < T_{2}$
B

$T_{1} > T_{2}$
C

$T_{1} = T_{2}$
D

$T_{1} = 2 T_{2}$

Solution

Using $x = Asinωt$
For $x = A / 2, {sinωT}_{1} = 1 / 2 \Rightarrow T_{1} = \frac{π}{6 ω}$
For $x = A, sinω (T_{1} + T_{2}) = 1 \Rightarrow T_{1} + T_{2} = \frac{π}{2 ω}$
$\Rightarrow T_{2} = \frac{π}{2 ω} - T_{1} = \frac{π}{2 ω} - \frac{π}{6 ω} = \frac{π}{3 ω} i . e . T_{1} < T_{2}$
Alternate method : In S.H.M., velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from 0 to $\frac{A}{2}$ will be less than the time taken to go from $\frac{A}{2}$ to A. Hence $T_{1} < T_{2} .$

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