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
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?
3 answers
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
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
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.
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.