Simple Harmonic Motion Solving for the phase and applying it.

[Wm_Davies]

Homework Statement

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?

Homework Equations

[tex]\omega=\sqrt{\frac{k}{m}}[/tex]
[tex]x=x_{m}cos(\omega t + \phi)[/tex]
[tex]v=-\omega x_{m}cos(\omega t + \phi)[/tex]

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 [tex]\phi=tan^{-1}(\frac{v}{x \omega})[/tex]

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

 

[rock.freak667]

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


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

 

[Wm_Davies]

Yes the formula for velocity was wrong. I divided the two which is how I got [tex]\phi[/tex] 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.

 

[rock.freak667]

Yes the formula for velocity was wrong. I divided the two which is how I got [tex]\phi[/tex] 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.

Could you please post your work on how you found all of your answers?

 

[Wm_Davies]

[tex]x=x_{m}cos(\omega t + \phi)[/tex]
[tex]v=-\omega x_{m}sin(\omega t + \phi)[/tex]

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


[tex]\omega t + \phi = tan^{-1}(\frac{v}{-x \omega})[/tex]

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 [tex]cos(\omega t + \phi)[/tex]

which came to... approximately 0.400912 meters

 

[rock.freak667]

What did you get for φ?

 

[Wm_Davies]

For [tex]\phi[/tex] I got approximately... 0.240112 radians

 

[rock.freak667]

Recheck it, I get -1.25 radians.

 

[Wm_Davies]

Yes I get -1.25897 radians when I solve for [tex]\omega t + \phi[/tex]