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
✓
Answer
(a)Using $x = A\sin \omega t$
For $x = A/2,\;\;\sin \omega {T_1} = 1/2 \Rightarrow {T_1} = \frac{\pi }{{6\omega }}$
For $x = A,\;\sin \omega ({T_1} + {T_2}) = 1 \Rightarrow {T_1} + {T_2} = \frac{\pi }{{2\omega }}$
$ \Rightarrow {T_2} = \frac{\pi }{{2\omega }} - {T_1} = \frac{\pi }{{2\omega }} - \frac{\pi }{{6\omega }} = \frac{\pi }{{3\omega }} $
For $x = A/2,\;\;\sin \omega {T_1} = 1/2 \Rightarrow {T_1} = \frac{\pi }{{6\omega }}$
For $x = A,\;\sin \omega ({T_1} + {T_2}) = 1 \Rightarrow {T_1} + {T_2} = \frac{\pi }{{2\omega }}$
$ \Rightarrow {T_2} = \frac{\pi }{{2\omega }} - {T_1} = \frac{\pi }{{2\omega }} - \frac{\pi }{{6\omega }} = \frac{\pi }{{3\omega }} $
$i.e.\;{T_1} < {T_2}$
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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