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

Option 1)
$T_{1} < T_{2}$

Option 2)
$T_{1} > T_{2}$

Option 3)
$T_{1} = T_{2}$

Option 4)
$T_{1} =2T_{2}$

Answers (1)

best_answer

Using $x = A\sin\omega t$

For $x = \frac{A}{2}$ , $\sin\omega T_{1} = \frac{1}{2} \Rightarrow T_{1} = \frac{\pi}{6\omega}$

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 $T_{1} < T_{2}$

Time Period -

Since all periodic motion repeat themselves in equal time interval. This minimum time interval is known as time period for oscillation.

- wherein

It is denoted by T.

Option 1)

$T_{1} < T_{2}$

This is correct.

Option 2)

$T_{1} > T_{2}$

This is incorrect.

Option 3)

$T_{1} = T_{2}$

This is incorrect.

Option 4)

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

This is incorrect.

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