A simple harmonic motion has an amplitude A and time period T. Find the time required by it to travel directly from $x = - \frac{A}{\sqrt{2}}$ , to $x = \frac{A}{\sqrt{2}}$ .

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Solution

Let the motion be $x = A sin (\frac{2 π}{T} t)$
Say, at any time $t_{1}$ particle is at $\frac{A}{\sqrt{2}}$
$\frac{A}{\sqrt{2}} = A sin (\frac{2 π}{T} t_{1})$
$\Rightarrow \frac{1}{\sqrt{2}} = sin (\frac{2 π}{T} t_{1})$
$\Rightarrow \frac{2 π}{T} t_{1} = \frac{π}{4}, \frac{3 π}{4} . . .$
$\Rightarrow t_{1} = \frac{T}{8}, \frac{3 T}{8} . . . e q n .1$
Now, at $t_{2}$ particle is at $- \frac{A}{\sqrt{2}}$
$\frac{A}{\sqrt{2}} = A sin (\frac{2 π}{T} t_{2})$
$\Rightarrow - \frac{1}{\sqrt{2}} = sin (\frac{2 π}{T} t_{2})$
$\Rightarrow \frac{2 π}{T} t_{2} = \frac{5 π}{4}, \frac{7 π}{4} . . . e q n .1$
$\Rightarrow t_{2} = \frac{5 T}{8}, \frac{7 T}{8} . . .$
From $e q n 1$ we can say that particle comes to $\frac{A}{\sqrt{2}}$ at time $\frac{T}{8}$ after starting from mean position and cross position and goes to extreme point. It again comes back to $\frac{A}{\sqrt{2}}$ at time $\frac{3 T}{8}$ . Then it crosses this position and further crossing mean position comes to $- \frac{A}{\sqrt{2}}$ at time $\frac{5 T}{8}$ . So, we can say that particle takes time $(\frac{5 T}{8} - \frac{3 T}{8}) = \frac{T}{4}$ in going from $\frac{A}{\sqrt{2}}$ to $- \frac{A}{\sqrt{2}}$ . And this should be equal to the time taken by the particle going from $- \frac{A}{\sqrt{2}}$ to $\frac{A}{\sqrt{2}}$

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