A particle executes SHM with periodic time of 6 seconds. The time taken for traversing a distance of half the amplitude from mean position is :

3 sec.

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2 sec.

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1 sec.

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1/2 sec.

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Solution

The correct option is D 1/2 sec.

Let eqn be $x = A s i n (w t)$

now $\frac{A}{2} = A s i n (w t)$

Where

$w = \frac{2 π}{T}$

From here $t = \frac{T}{12}$
$t = \frac{6}{12} = \frac{1}{2} s e c .$

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A particle executes SHM with periodic time of 6 seconds. The time taken for traversing a distance of half the amplitude from mean position is :

The correct option is D 1/2 sec. Let eqn be x=Asin(wt) now A2=Asin(wt) Where w=2πT From here t=T12 t=612=12sec.

The correct option is D 1/2 sec.

Let eqn be $x = A s i n (w t)$

now $\frac{A}{2} = A s i n (w t)$

Where

$w = \frac{2 π}{T}$

From here $t = \frac{T}{12}$
$t = \frac{6}{12} = \frac{1}{2} s e c .$