Question

An object executes simple harmonic motion with an amplitude A. (Use any variable or symbol stated above as necessary.) (a) At what values of its position does its speed equal half its maximum speed? x = ± (b) At what values of its position does its potential energy equal half the total energy?

Answer 1

Answer:

(a) x=ASin(ωt+Ф₀)=±(√3)A/2

(b) x=±(√2)A/2

Explanation:

For part (a)

V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)

Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3

x=ASin(ωt+Ф₀)=±(√3)A/2

For part(b)

U=0.5E and U+K=E→K=0.5E

E=K(Max)

(1/2)mv²=(0.5)(1/2)m(Vmax)²

V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4

x=±(√2)A/2