An object with mass 0.250 kg is acted on by an elastic restoring force with forc
ID: 1504067 • Letter: A
Question
An object with mass 0.250 kg is acted on by an elastic restoring force with force constant 10.0 N/m . The object is set into oscillation with an initial potential energy of 0.150 J and an initial kinetic energy of 6.60×102 J .
What is the amplitude of oscillation?
What is the potential energy when the displacement is one-half the amplitude?
At what displacement are the kinetic and potential energies equal?
What is the value of the phase angle if the initial velocity is positive and the initial displacement is negative?
Explanation / Answer
Total energy = PE + KE = kA^2 / 2
0.150 + 0.0660 = 10 A^2 /2
A = 0.208 m
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when x = A/2 = 0.104 m
PE = 10 (0.104)^2 / 2 = 0.054 J
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total energy will not change .
when PE = KE
then PE = KE = (0.150 + 0.066)/2 = 0.108
PE = 10 x^2 /2 = 0.108
x = 0.147 m
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x = A sin(wt + phi )
at t =0
x = A sin(phi)
for inital position, 0.150 = 10 x^2 /2
x = 0.173 m
- 0.173 = 0.208 sin(phi)
phi = - 0.313 pi Or - 56.4 deg