Question
A mass is oscillating with amplitude A at the end of a spring.
How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?
How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals the kinetic energy?
Answers
Elena
x=A•sinωt
v =dx/dt = A•ω•cosωt
PE = KE
kx^2/2 = mv^2/2
k = m•ω^2
m•ω^2• (A•sinωt)^2/2 = m•( A•ω•cosωt)^2/2
(sinωt)^2 = (cosωt)^2
(tan ωt)^2 = 1
ωt = π/4
x=A•sinωt = A•sin(π/4) =0.707•A
v =dx/dt = A•ω•cosωt
PE = KE
kx^2/2 = mv^2/2
k = m•ω^2
m•ω^2• (A•sinωt)^2/2 = m•( A•ω•cosωt)^2/2
(sinωt)^2 = (cosωt)^2
(tan ωt)^2 = 1
ωt = π/4
x=A•sinωt = A•sin(π/4) =0.707•A
Anonymous
It is not clear