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

T_{1} < T_{2}

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T_{1} > T_{2}

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T_{1} = T_{2}

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T_{1} = 2 T_{2}

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Solution

The correct option is A $T_{1} < T_{2}$
Using $x = A sin ω t$
For $x = \frac{A}{2}, sin ω T_{1} = \frac{1}{2} \Rightarrow T_{1} = \frac{π}{6 ω}$
For $x = A, sin ω (T_{1} + T_{2}) = 1 \Rightarrow T_{1} + T_{2} = \frac{π}{2 ω}$
$\Rightarrow T_{2} = \frac{π}{2 ω} - T_{1} = \frac{π}{2 ω} - \frac{π}{6 ω} = \frac{π}{3 ω} i.e. T_{1} < T_{2}$

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