# Finding the phase angle in simple harmonic motion

1. Oct 25, 2005

### erik-the-red

Question:

A frictionless block of mass 2.35 kg is attached to an ideal spring with force constant 310 N/m. At t=0 the spring is neither stretched nor compressed and the block is moving in the negative direction at a speed of 12.6 m/s.

1. Find the amplitude.

I got this without a problem. I used the formula A = square root of $$x_i^2$$ + $$v_i^2$$/$$\omega^2$$. $$\omega = \sqrt{k/m} = 11.49$$ rad/s. Plugging in known values results in 1.10 m, which is correct.

2. Find the phase angle.

Here's where I don't know why my answer is not correct.

I use the equation $$v_i = -\omega * A * sin(\omega*t + \phi)$$. I know the initial velocity, I know the angular frequency, and I know the amplitude. I'm solving for the angle in radians.

$$-12.6 = -(11.49)(1.10)(sin(\phi)$$ This is at time t, so $$\omega * t$$ = 0.

I get $$\phi = 85.5 ^\circ$$. Converting it into radians, it's 1.49 (rad).

This isn't right, though.

Last edited: Oct 25, 2005
2. Oct 25, 2005

### Staff: Mentor

If your numbers were computed a bit more accurately, you'd get:
$$-12.6 = - (12.6) \sin \phi$$, or
$$1 = \sin \phi$$, thus $$\phi = \pi /2$$ (radians)

Last edited: Oct 25, 2005
3. Oct 25, 2005

### erik-the-red

I'm solving for a phase angle here, $$\phi$$. What additive constant could possibly exist?

4. Oct 25, 2005