A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is:
A) zero
B) 12.5 J
C) 25 J
D) 37.5 J
E) 50 J

Question

Quizplus · Accepted Answer

The total energy of the spring block system is the sum of its potential energy and kinetic energy:
E = 1/2 kA^2 + 1/2 mv^2
where k is the spring constant, A is the amplitude, and v is the velocity of the block.
At half the amplitude, the displacement x = A/2. The potential energy at this point is:
U = 1/2 k (A/2)^2 = 1/8 kA^2
Since the total energy is 50 J, the kinetic energy at this point is:
K = E - U = 50 - 1/8 kA^2
We don't know the values of k and A, but we can still use the information given in the question to find the answer. Since the motion is simple harmonic, the velocity and displacement are related by:
v^2 = k/m (A^2 - x^2)
At x = A/2,
v^2 = k/m (A^2 - A^2/4) = 3/4 kA^2/m
The kinetic energy at this point is therefore:
K = 1/2 mv^2 = 1/2 (3/4 kA^2/m) = 3/8 kA^2
Substituting this into the expression for K above, we get:
3/8 kA^2 = 50 - 1/8 kA^2
Solving for kA^2, we get:
kA^2 = 200/5 = 40
Therefore, the kinetic energy at half the amplitude is:
K = 3/8 kA^2 = 3/8 * 40 = 15 J
Since the answer choices are in increments of 12.5 J, the closest option is D: 37.5 J.

A Block Attached to a Spring Undergoes Simple Harmonic Motion