Finding the phase angle in simple harmonic motion

[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.

 

[Doc Al]

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)

 

[erik-the-red]

My response is, "My answer is off by an additive constant."

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

 

[Doc Al]

I messed up the sign in my previous post (D'oh!); your first answer was close. (See my correction.)

 

[erik-the-red]

Thanks. My idea of direction was screwed up, but it didn't matter for the first part of the question because velocity is squared.

Round once too many and the error becomes greater than two percent. Argh.