A particle executes simple harmonic motion between $x = - A$ and $x = + A$ . The time taken for it to go from the mean position $O$ to $A / 2$ is $T_{1}$ and to go from $A / 2$ 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 $T_{1} < T_{2}$

We know that velocity of the particle in simple harmonic motion is $v = ω \sqrt{A^{2} - x^{2}}$ where $ω$ is angular speed, $A$ is amplitude and $x$ is displacement from mean position.
So, when $x$ increases, the velocity $v$ decrease.

Therefore, particle will take more time to go from $x = \frac{A}{2}$ than to go from $x = 0$ to $x = \frac{A}{2}$ .
Hence, option (a) is correct.

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