# Simple Harmonic Motion Solving for the phase and applying it.

1. Apr 8, 2010

### Wm_Davies

1. The problem statement, all variables and given/known data
I am having a hard time solving this problem and this is my last chance to solve it before it is due and I have to do this for a test tomorrow.

A simple harmonic oscillator consists of a block of mass 3.90 kg attached to a spring of spring constant 330 N/m. When t = 0.570 s, the position and velocity of the block are x = 0.123 m and v = 3.510 m/s. (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at t = 0 s?

2. Relevant equations

$$\omega=\sqrt{\frac{k}{m}}$$
$$x=x_{m}cos(\omega t + \phi)$$
$$v=-\omega x_{m}cos(\omega t + \phi)$$

3. The attempt at a solution

Okay so I tried to solve for the phase constant by dividing the velocity function by the position function.

This gave me the result of $$\phi=tan^{-1}(\frac{v}{x \omega})$$

I plugged in $$\phi$$ to both the position functions and my answers are wrong. So I am not sure as to what I need to do or why.

Last edited: Apr 8, 2010
2. Apr 8, 2010

### rock.freak667

Your expression for v should be v=-ωxmsin(ωt+φ)

Also consider what v/x gives and then substitute the conditions to get φ.

3. Apr 8, 2010

### Wm_Davies

Yes the formula for velocity was wrong. I divided the two which is how I got $$\phi$$ is inverse tangen times v divided by the quantity of position times the angular frequency. I must be doing something right because I am able to solve for the amplitude of the function, but I cannot get the position at t=0 and the velocity at t=0.

4. Apr 8, 2010

5. Apr 8, 2010

### Wm_Davies

$$x=x_{m}cos(\omega t + \phi)$$
$$v=-\omega x_{m}sin(\omega t + \phi)$$

So I divided these two functions which gave me the result of......

$$\omega t + \phi = tan^{-1}(\frac{v}{-x \omega})$$

Since this is equal to all of the stuff in the position function and the velocity function I put it into the those functions after solving for the amplitude. This cannot lead to a correct answer because the velocity and position in the above mentioned expression are not at time = 0,

P.S. I solved for the amplitude by dividing x at t=.570s by the $$cos(\omega t + \phi)$$

which came to...... approximately 0.400912 meters

6. Apr 8, 2010

### rock.freak667

What did you get for φ?

7. Apr 8, 2010

### Wm_Davies

For $$\phi$$ I got approximately.... 0.240112 radians

8. Apr 8, 2010

### rock.freak667

Recheck it, I get -1.25 radians.

9. Apr 8, 2010

### Wm_Davies

Yes I get -1.25897 radians when I solve for $$\omega t + \phi$$