Question

A simple harmonlc oscillator has a period of 0.01 sec and an amplitude of 0.2 m. The
magnitude of velocity in m/s at the centre of oscillation is:
a) 100
b) 100pi
c) 20pi
d)40pi​

Answer 1

Answer:

answer is 40 pi

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

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Answer 2

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=ω2π  ⇒ω=200π

∴ Velocity=0.2×200π=40π