Question

A particle moving with simple harmonic motion has a period 0.01 sec and amplitude 0.04 m. Find the acceleration when it is 0.003 m away from the mean position and maximum velocity

Answer 1

Answer:

40π

Explanation:

Let equation of SHM is x=Asin(ωt), Where A=0.2 m,

∴v=Aωcos(ωt),  

∴ Velocity at mean position v=Aω

Given, Time Period=0.01=  

ω

2π

​

  ⇒ω=200π

∴ Velocity=0.2×200π=40π

Answer 2

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