A particle in SHM has a period of $4$ sec. It takes time $t_{1}$ to start from mean position and reach half the amplitude in another case it takes a time $t_{2}$ to start from extreme position and reach half the amplitude, then

\frac{t_{1}}{t_{2}}

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\frac{t_{1}}{t_{2}}

\frac{1}{2}

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\frac{t_{1}}{t_{2}}

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\frac{t_{1}}{t_{2}}

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Solution

The correct option is B $\frac{t_{1}}{t_{2}}$ = $\frac{1}{2}$

SHM can be described by the equation $x = A s i n (\frac{2 π t}{T})$

$t = 0, x = 0$

$x = A / 2$ at time t = $\frac{2 π t}{T} = \frac{π}{6}$ => $t = \frac{T}{12}$

$x = A$ at time t = $\frac{2 π t}{T} = \frac{π}{2}$ => $t = \frac{T}{4}$

So time to move from mean position to half of amplitude, $t_{1} = \frac{T}{12}$ s

Time to move from extreme to half of amplitude = time to move from half of amplitude to extreme $= t_{2} = \frac{T}{4} - \frac{T}{12} = \frac{T}{6}$

Therefore $\frac{t_{1}}{t_{2}} = \frac{1}{2}$

Answer. B

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