Question

A body moving in simple harmonic motion has an amplitude of 10cm and a frequency of 100Hz. Find i.mthe period of oscillation, ii. the acceleration at the maximum displacement, iii. the velocity at the centre of motion​

Answer 1

Explanation:

= 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

When displacement is y, then acceleration, a=−ω2y

Velocity, V=ωr2−y2

Case (a)    When y=5cm=0.05m

a=−(10π)2×0.05=−5π2m/s2

V=10π×(0.05)2−(0.05)2=0

Case (b)    When y=3cm=0.03m

a=−(10π)2×0.03=−3π2m/s2

V=10π×(0.05)2−(0.03)2=10π×0.04=0.4π m/s

Case (c)    When y=0

a=−(10π)2×0=0

V=10π×(0.05)2−02=

hope it helps. pls follow and mark as brainliest