Question

A 96.3 g mass is attached to a horizontal spring with a spring constant of 1.04 N/m and released from rest with an amplitude of 27.1 cm.
What is the velocity of the mass when it is halfway to the equilibrium position if the surface is frictionless?

Answer 1

the KE at the null position is equal to the starting PE

1/2 m v^2=1/2 k x^2

solve for v

Answer 2

If the amplitude is A, you are required
to find the velocity of the mass when
it is at a distance A/2 from the equilibrium position (i.e. midway between the extreme and equilibrium positions)

use the expression:

v = w*sqrt(A^2 - x^2)
where v - velocity at displacement x
x - displacement from the eq. position
w - oscillation freq. in rad/sec
= sqrt(k/m) = sqrt(1.04/.0963)
A - Amplitude = 0.271m

So v = sqrt(1.04/.0963)*sqrt[A^2 -(A/2)^2]
v = 0.77 m/s or 77cm/s

Answer 3

The equation of oscillation is
x=A•cos(ωt) (as at t=0 x=A),
velocity of the mass is
v =dx/dt = - A•ω•sin(ωt).
If x =A/2, then
A/2=A•cos(ωt),
cos(ωt) =0.5 => sin(ωt)=sqrt(1- cos²(ωt)) =0.866,
ω=sqrt(k/m) =sqrt(1.04/0.0963) =3.29 rad/s,
v = - A•ω•sin(ωt) =-0.271•3.29•0.866 = - 0.77 m/s.