Question

A 492g oscillator has a speed of 99.5 cm/s when its displacement is 3.60 cm and 72.5 cm/s when its displacement is 5.10 cm.
What is the oscillator's maximum speed?

I'm suck with this problem, can someone please go through it with me step by step, please and thanks

Answer 1

Total energy (kinetic plus potential energy) at any point in the oscillator is a constant.

KE = (1/2)m v^2
PE = (1/2)kx^2

TE = KE + PE
KE1 + PE1 = KE2 + PE2
0.2435J + PE(3.60cm) = 0.1293J + PE(5.10cm)
0.1142J = PE(5.10cm) - PE(3.60cm)
0.1142J = (1/2) k(5.10cm)^2 - (1/2) k (3.60cm)^2
0.1142J = (1/2) k ((5.10cm)^2 - (3.60cm)^2))
0.2284J = k (13.05cm^2)
k = 175.0 kg/s^2

PE(3.60cm) = (1/2) (175.0kg/s^2) (3.60cm)^2
PE(3.60cm) = 0.1134J

TE = KE(3.60cm) + PE(3.60cm)
TE = 0.2435J + 0.1134J
TE = 0.3569J

At maximum speed, all the energy is kinetic.
KE = TE
(1/2) m v^2 = TE
v^2 = 2 TE / m
v = sqrt(2 TE / m)
v = sqrt( 2 0.3569J / 492g)
v = sqrt(1.451 m^2/s^2)
v = 120cm/s