The displacement of a block of mass 778g attached to a spring whose spring constant is 38N/m is given by x=Asin(ωt) where A=25cm. In the first complete cycle find the values of x and t at which the kinetic energy is equal to one half the potential energy.

Question

First position:
First time:

Second position:
Second time:

Third position:
Third time:

Fourth position:
Fourth time:

time in s and position in cm please

Answer 1

forget cm, doing seconds and meters - SCI units

Pe = (1/2) k |x|^2
Ke = (1/2) m |v|^2

so for Ke = 1/2 Pe
m |v|^2 = .5 k |x|^2
but ω^2 = k/m
so
|v|^2 = .5 ω^2 |x|^2 At our spots
[A ω cos ωt]^2 = .5ω^2[A sin ωt]^2
A^2 ω^2 cos^2ωt = .5ω^2 A^2sin^2wt
cos^2 ωt = .5 sin^2 ωt
cos^2 ωt/sin^2 ωt = .5

cos/sin = + or - sqrt(1/2)