the phase of the spring oscillator when it is first 2 cm away from the equilibrium position is:

Question

If a horizontally placed spring oscillator pulls 4 cm positive along the X axis and then releases, the phase of the spring oscillator when it is first 2 cm away from the equilibrium position is:

A. -π

B 0

C -π/3

D π/3

Answer 1

Solution:

The general formula for displacement of a horizontally stretched spring is X = X_m cos( ωt+φ) where ( ωt+φ) is the phase angle under the cosine curve. When the spring is compressed 4 cm to the left from its equilibrium position, that denotes the value of X_m or the initial position as 4 cm positive. When the spring is released and reaches 2 cm from the equilibrium point. That point is the value of X, the final position of the spring mass or 2 cm positive. Note that both values are +ve positioned left of the equilibrium point in the compressed condition.

Solving for cos( ωt+φ) = X / X_m = 2cm / 4cm = 1/2 = 0.5 , then cos^-1cos( ωt+φ) = ( ωt+φ) = cos^-1( 0.5) = 60º = π/3

Answer is D

Note that if the spring is outstretched to the right of the equilibrium point, the values of X and X_m would both be negative leading to the same phase value. However if the required phase is from an compressed position to an outstretched position or vice versa, the value of the phase angle would be much larger. Try it for yourself as an exercise.

The sign value has to do with the direction of the force. When the spring is compressed, the stored energy in the spring will cause the force to be directed right, hence positive direction of the force vector, and when the spring is outstretched, the force would be directed left, hence negative direction of the force vector.