A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ________ cm/s.

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Solution

Q.
A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ________ cm/s. [2024]

Ans.
(12) $Given : A = 0.06 cm$

$T = \frac{2 π}{ω} = 3.14 \Rightarrow ω = 2 rad / s$

$V_{\max} = ω A = 2 \times \frac{6}{100} = 12 cm / s$

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