# Simple harmonic motion problem help.

1. Oct 13, 2012

### Eats Dirt

1. The problem statement, all variables and given/known data

particle experiencing SHM with frequency f= 10 hz
find the displacement x at any time t for the following initial conditions.
@ t=0 x=0.25m v=0.1 m/s

2. Relevant equations

x=Asin(ωt+∅)
v=Aωcos(ωt+∅)

3. The attempt at a solution

So with frequency I find ω which then is subbed into the x=Asin(ωt+∅) @ t=0 to yield 0.25=Asin(∅).

I then get v=Aωcos(ωt+∅) and sub in for v and t and ω and get 0.25=A(20∏)cos(∅)
rearrange:

0.25/(20∏)=Acos(∅)

I then divide these equations by each other and rearrange to get:

arctan(50∏)= ∅

then with this I go back to 0.25=Asin(∅) and sub in ∅ rearrange to solve for A and its wrong! I get x=(0.16)sin(20∏t+1.56).

answer in book says x=0.25cos(20∏t)+0.00159sin(20∏t). it says also acceptable solutions would be x=0.25sin(20∏t+1.56) and its cos variant.

can someone please help I am confused about how they got this answer and how it is in that form.

2. Oct 13, 2012

### Spinnor

I like the books answers, I think you made a math error.

Edit, I like the books first answer, still can't get the books second expression to work?

The 1.56 in the books answer should be 1.5644?

Last edited: Oct 13, 2012
3. Oct 13, 2012

### Eats Dirt

hurmm but why would Amplitude be 0.25 when it has a velocity that is not 0 meaning it is not at its maximum or minimum? and yes in the book they round.

4. Oct 13, 2012

### Spinnor

Because they give you the initial conditions at some point between max and min. The books answers (at least the first one) work when you properly substitute t = 0.

5. Oct 13, 2012

### frogjg2003

While $x(t)=A\sin(\omega t+\phi)$ is a correct general solution to the harmonic oscillator, $x(t)=A\cos(\omega t)+B\sin(\omega t)$ is also correct and has the added bonus that $x(0)=A$ and $x'(0)=B\omega$.

Edit:
You can go from one to the other by using the identitiy:
$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$
and going from $x(t)=B\cos(\omega t+\phi)$ can be done with
$$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$

Last edited: Oct 13, 2012
6. Oct 13, 2012

### frogjg2003

The reason you can't get the books answer is because you used .25m for the velocity instead of .1m/s.

7. Oct 14, 2012

### Eats Dirt

thank you all this is more clear now!