college physics

Question

When a mass of m = 304g is attached to a spring and the mass spring system is set into oscillatory motion, the period of the motion is r = 0.467s Determine the amplitude (meter) of the oscillation, if the total energy of the oscillating system is 0.163 J

Answer 1

Work with the total energy:

E = kA²/2

k = mω² ⇒

E = mω²A²/2 ⇒

A = √[2E / (mω²)]

ω = 2π/T ⇒

A = [2E / m(2π/T)²]^1/2

You can simplify this a little more if you wish, and plug in

E = 0.163 J

m = 0.304 kg

T = 0.467 s

and get A in meters.