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 $x = A$ .

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Solution

Let the motion be $x = A sin (\frac{2 π}{T} t)$
At $t = t_{1}$ , $x = \frac{A}{\sqrt{2}}$
$\frac{A}{\sqrt{2}} = A sin (\frac{2 π}{T} t_{1})$
$\Rightarrow \frac{2 π}{T} t_{1} = \frac{π}{4}$
$\Rightarrow t_{1} = \frac{T}{8}$
At $t_{2}$ , $x = A$ So, $A = A sin (\frac{2 π}{T} t_{2})$
$\Rightarrow \frac{2 π}{T} t_{2} = \frac{π}{2}$
$\Rightarrow t_{2} = \frac{T}{4}$
So, the particle will take time $(t_{2} - t_{1})$
$= (\frac{T}{4} - \frac{T}{8}) = \frac{T}{8} s e c$ in going from $\frac{A}{\sqrt{2}}$ to $A$

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