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Question

A particle is executing S.H.M. between  $$ x = \pm A. $$ The time taken to go from 0 to $$ \frac {A}{2} $$ is $$ T_1 $$ and to go from $$ \frac {A}{2} $$ to A is $$ T_2 $$ then:


A
T1<T2
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B
T1>T2
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C
T1=T2
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D
T1=2T2
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Solution

The correct option is A $$ T_1 < T_2 $$
let $$t=0, x=0$$
Then, $$x=A\sin (wt)$$
At $$t=T_1, x=\dfrac{A}{2}$$
$$\Rightarrow \dfrac{A}{2}=A\sin (wT_1)$$
$$\Rightarrow  \sin wt_1=1/2$$
$$\Rightarrow T_1=\dfrac{\pi}{6w}$$
At $$t=(T_1+T_2), x=A$$
$$\Rightarrow  A=A\sin (w[T_1+T_2])$$
$$\Rightarrow \sin [w(T_1+T_2)]=1$$
$$T_1+T_2=\dfrac{\pi}{2w}\Rightarrow T_2=\dfrac{\pi}{2w}-T_1$$
$$\Rightarrow T_2=\dfrac{\pi}{2w}-\dfrac{\pi}{6w}=\dfrac{\pi}{3w}$$
So, $$\dfrac{\pi}{3w}=2\left(\dfrac{\pi}{6w}\right)$$
$$\Rightarrow  T_2=2T_1 \Rightarrow T_2>T_1$$
Option- $$A$$ is correct.

Physics

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