Help I get exactly 1/2 of the correct answer for phase constant calculation

[996gt2]

Homework Statement

An air-track glider attached to a spring oscillates with a period of 1.50 s. At t=0s, the glider is 5.60 cm left of the equilibrium position and moving to the right at 40.6 cm/s.

What is the phase constant?

2. Relevant solutions

The Attempt at a Solution

$$T=1.50s$$
$$\omega=2\pi/T=4.189$$

$$1: A cos \phi =-0.056m$$
$$2: -\omega A sin \phi =0.406m/s$$
$$-A sin \phi=0.406/4.189=0.097$$Divide equation 2 by equation 1:

$$- tan \phi=0.097/-0.056$$
$$\phi=1.05 rad$$

However, the correct answer is exactly twice my answer, and also negative of what I got. Can anyone explain what dumb mistake I'm making here? Thanks!

 

[alphysicist]

Homework Statement

An air-track glider attached to a spring oscillates with a period of 1.50 s. At t=0s, the glider is 5.60 cm left of the equilibrium position and moving to the right at 40.6 cm/s.

What is the phase constant?

2. Relevant solutions

The Attempt at a Solution

$$T=1.50s$$
$$\omega=2\pi/T=4.189$$

$$1: A cos \phi =-0.056m$$
$$2: -\omega A sin \phi =0.406m/s$$
$$-A sin \phi=0.406/4.189=0.097$$


Divide equation 2 by equation 1:

$$- tan \phi=0.097/-0.056$$
$$\phi=1.05 rad$$

However, the correct answer is exactly twice my answer, and also negative of what I got. Can anyone explain what dumb mistake I'm making here? Thanks!

When you take the inverse tangent, you have to check whether the answer the calculator gives is the correct answer or if the correct answer is 180 degrees away. (The problem is that the calculator does not discriminate between:

$$\tan^{-1}\left( \frac{-0.097}{-0.056} \right) \mbox{ and } \tan^{-1}\left( \frac{0.097}{0.056} \right)$$
even though when we write them like that we think of one being in the first quadrant and one being in the third quadrant.

So you got an answer of 1.05rad. Now you have to check whether that angle, or (1.05 - $\pi$)rad is the right answer. So plug them both into your original sine and cosine relations and see which angle gives the correct signs for both.

(Instead of checking (1.05 - $\pi$)rad, you could use (1.05 + $\pi$)rad; they are the same angle.)

 

[996gt2]

Ah ok thanks very much for the tip!