Physics: Simple Harmonic Motion

**Problem: **A mass is attached to the end of a spring and set into simple harmonic motion with an amplitude A on a horizontal frictionless surface. Determine the following in terms of only the variable A.

(a) Magnitude of the position (in terms of A) of the oscillating mass when its speed is 50% of its maximum value.

Solution:

Using the conservation of energy formula, solve for Vmax in terms of P, M, K, and A (where P is the ratio).

Etot = KE + PE = K(A²)/2 = MV²/2 + KX²/2

We know that KE is maximized when the velocity is at its maximum and that KE is at its maximum when KE = Etotal, and thus PE = 0.

Etot = KE = KA²/2 = MVmax²/2

Solving for Vmax, we get:

Vmax = sqrt(KA²/M)

To get Vmax down to 50% of its value, we simply refer to it as PVmax, where P = .50      PVmax = P * sqrt(KA²/M)

Now, plug the formula we just found for PVmax into the conservation of energy formula from above to get:

X² = A²(1-P²)

Answer: X = 0.866A

(b) Magnitude of the position (in terms of A) of the oscillating mass when the elastic potential energy of the spring is 50% of the total energy of the oscillating system.

**Solution: **

Similarly, here we just set PE = (P)Etot, where P is the given ratio

PE = (P)Etot = KX²/2 = PKA²/2

Solving for X, we get:

X² = PA²

**Answer: **X = 0.707A