Challenging physics question regarding amplitude of oscillation

Question

A 150 gram ball attached to a spring with spring constant

2.30 N/m oscillates horizontally on a frictionless table. Its velocity is 20.0 cm/s when x= -5.00 cm.

What is the amplitude of oscillation in cm?

What is the speed of the ball when x= 3.00 cm? answer in cm/s

Answer 1

Let:

w be the angular frequency,

k be the spring constant,

m be the mass,

a be the amplitude,

v be the speed,

x be the displacement from equilibrium.

w^2 = k / m

v = w sqrt(a^2 - x^2)

v = sqrt[ k(a^2 - x^2) / m] ...(1)

v^2 = k(a^2 - x^2) / m

a^2 = mv^2 / k + x^2 ...(2)

From (2) when x = 5.00 cm:

a = sqrt[ 0.150 * 0.0200^2 / 2.30 + 0.05^2 ]

= 0.0503 m

= 5.03 cm.

From (1) when x = 3.00 cm:

v = sqrt[ (2.30 / 0.15)(0.0503^2 - 0.03^2) ]

= 0.158 m/s

= 15.8 cm/s.