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?
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
To find the velocity of the mass when it is halfway to the equilibrium position, we need to apply the principles of simple harmonic motion.
1. Determine the equilibrium position: The equilibrium position is the position where the mass is at rest. Since the spring is stretched to an amplitude of 27.1 cm, the equilibrium position is halfway between the maximum displacement (27.1 cm) and the minimum displacement (-27.1 cm), which is at 0 cm.
2. Calculate the potential energy at the equilibrium position: The potential energy at the equilibrium position is zero since the mass is at rest, and there is no additional spring potential energy.
3. Use the principle of conservation of mechanical energy: At the highest point of the motion (maximum displacement), all the potential energy is converted into kinetic energy. At the equilibrium position (halfway), half of the potential energy is converted into kinetic energy. Therefore, the change in potential energy is equal to the change in kinetic energy.
4. Calculate the potential energy at the maximum displacement:
Potential Energy = (1/2) * k * x^2
where k is the spring constant and x is the maximum displacement from the equilibrium position.
Potential Energy = (1/2) * 1.04 N/m * (27.1 cm)^2
5. Calculate the potential energy at the equilibrium position (halfway):
Potential Energy at Equilibrium Position = (1/2) * k * x^2
where x is the displacement from the equilibrium position.
Potential Energy at Equilibrium Position = (1/2) * 1.04 N/m * (27.1 cm/2)^2
6. Calculate the change in potential energy:
Change in Potential Energy = Potential Energy at Maximum Displacement - Potential Energy at Equilibrium Position
7. Calculate the change in kinetic energy:
Change in Kinetic Energy = Change in Potential Energy
8. Use the equation for kinetic energy:
Kinetic Energy = (1/2) * m * v^2
where m is the mass and v is the velocity.
9. Set the change in kinetic energy equal to the change in potential energy:
(1/2) * m * v^2 = Change in Potential Energy
10. Solve for v:
v = sqrt((2 * Change in Potential Energy) / m)
Now you can plug in the values and calculate the velocity of the mass when it is halfway to the equilibrium position.
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
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.