Simple harmonic motion problem help.

[Eats Dirt]

Homework Statement

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

Homework Equations

x=Asin(ωt+∅)
v=Aωcos(ωt+∅)

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.

 

[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?

 

[Eats Dirt]

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?

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.

 

[Spinnor]

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.

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.

 

[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)

 

[frogjg2003]

@ t=0 x=0.25m v=0.1 m/s
...
I then get v=Aωcos(ωt+∅) and sub in for v and t and ω and get 0.25=A(20∏)cos(∅)

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

 

[Eats Dirt]

thank you all this is more clear now!