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>>Oscillations
>>Simple Harmonic Motion
>> $The displacement of a particle executing$

Question

The displacement of a particle executing SHM at any instant t is $x = 0.01 sin 100 (t + 0.05)$ then its time period will be

A

0.06 s

B

0.2 s

C

0.1 s

D

0.02 s

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Updated on : 2022-09-05

Solution

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Correct option is A)

$x = 0.01 sin (100 t + 5)$
We have $x = A sin (ω t + θ)$
On comparing we get
Thus, here $ω = 100$
Thus, $T = \frac{2 π}{ω} = \frac{2 π}{1 0 0} = 0.06 seconds .$

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