Question

A simple harmonic oscillator has a period of 0.01 s and an amplitude of 0.2 m. The magnitude of the velocity in ms-1 at the centre of oscillation is​

Answer 1

Answer:

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

At the centre of oscillation means at mean position. velocity at mean position is the maximum whose value is Aw where A is amplitude and w is angular velocity.

Given, time period (T) = 0.01

we know,

w = 2π/T

=) w = 2π/0.01

=) w = 200π

Given A = 0.2

=) Velocity = Aw

=) velocity = 0.2 × 200π

=) velocity = 40π m/s (answer)

Hope this helps you ☺️