Energy of Harmonic Oscillators

Energy of Harmonic Oscillators
Part A = A
Part B = A
Part C = moving toward equilibrium.
Part D = C
Part E = C
Part F = D
Part G = 3/8kA²
Solutions Below:

Consider a harmonic oscillator at four different moments, labeled A, B, C, and D, as shown in the figure . Assume that the force constant k, the mass of the block, m, and the amplitude of vibrations, A, are given. Answer the following questions:

Part A

Which moment corresponds to the maximum potential energy of the system?

A
B
C
D
The maximum PE is when the spring is fully compressed. D might look like the right answer, but actually A is when the spring is at amplitude. Even though it’s stretched, that still counts as compression:
A


Part B

Which moment corresponds to the minimum kinetic energy of the system?

A
B
C
D
When PE is maximized, KE is minimized, so the correct answer is the same as from Part A:
A


Part C

Consider the block in the process of oscillating. If the kinetic energy of the block is increasing, the block must be:

A. at the equilibrium position.
B. at the amplitude displacement.
C. moving to the right.
D. moving to the left.
E. moving away from equilibrium.
F. moving toward equilibrium.
KE is maximized when PE is minimized. PE is greatest at amplitude and KE is greatest at equilibrium. So choice F is correct:
F. moving toward equilibrium.


Part D

Which moment corresponds to the maximum kinetic energy of the system?

A
B
C
D
KE is greatest at equilibrium:
C


Part E

Which moment corresponds to the minimum potential energy of the system?

A
B
C
D
Minimum PE is when KE is greatest, i.e. at equilibrium:
C


Part F

At which moment is K = U?

When U = KE, U = 1/2U_max (this is just a fact, which we won’t bother solving for here). So:
U = 1/2(U)
1/2kx² = 1/2(1/2kA²)
x² = 1/2A²
x = sqrt(1/2A²)
x = Asqrt(2)/2
Note- if it isn’t obvious where the sqrt(2)/2 came from, use 2/4 for the fraction above instead of 1/2:
x = sqrt((2/4)A²)
x = Asqrt(2)/2
The diagram doesn’t give this as an answer, but it does give -Asqrt(2)/2, which is equivalent:
D


Part G

Find the kinetic energy K of the block at the moment labeled B.
Express your answer in terms of k and A.

Since total energy = KE + PE and we only have enough information to find PE, we can work backwards by first finding the maximum PE (the PE when the spring is fully compressed) and then subtracting the PE at point B.
Maximum PE:
PE_max = 1/2kA²
PE at Point B:
PE_B = 1/2k(A/2)²
PE_B = 1/2k(A²/4)
PE_B = 1/8k(A²)
Now find the difference between the PEs, and that difference must be KE since there is no friction:
KE_B = PE_max – PE_B
KE_B = 1/2kA² – 1/8k(A²)

KE_B = 3/8kA²