Question

The maximum displacement of an oscillating particle is 0.05 . If it's time period is 1.57s.what is the velocity at the mean position and what is it's acceleration at the extreme position

Answer 1

$Answer =>$

The velocity at the mean position will be 0.2 m/s

$Explanation =>$

Given Time period, T = 1.57s

we know that,

angular frequency, ω = 2π/T

=> ω = 2 x 3.14/1.57 = 4

The maximum displacement = a = 0.05 m

The velocity will be maximum at the mean position, Hence velocity at the mean position,

Vmax = aω = 0.05 x 4 = 0.2 m/s

Hence velocity at the mean position will be 0.2 m/s

Answer 2

Answer:

0.2 m/s is the answer...