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. $T_1<T_2$
• B. $T_1>T_2$
• C. $T_1=T_2$
• D. $T_1=2T_2$

Answer 1

The correct option is A $T_1<T_2$

let $t=0,x=0$

Then, $x=A\sin \left(wt\right)$

At $t=T_1,x=\frac{A}{2}$

$\Rightarrow \frac{A}{2}=A\sin \left(w{T}_1\right)$

$\Rightarrow \sin w{t}_1=1/2$

$\Rightarrow T_1=\frac{\pi }{6w}$

At $t=(T_1+T_2),x=A$

$\Rightarrow A=A\sin \left(w\right[T_1+T_2\left]\right)$

$\Rightarrow \sin \left[w\right(T_1+T_2\left)\right]=1$

$T_1+T_2=\frac{\pi }{2w}\Rightarrow T_2=\frac{\pi }{2w}-T_1$

$\Rightarrow T_2=\frac{\pi }{2w}-\frac{\pi }{6w}=\frac{\pi }{3w}$

So, $\frac{\pi }{3w}=2\left(\frac{\pi }{6w}\right)$

$\Rightarrow T_2=2T_1\Rightarrow T_2>T_1$

Option- $A$ is correct.