# An object goes through simple harmonic motion with its position described as

1. Nov 25, 2008

### losthelp

An object goes through simple harmonic motion with its position described as....

1. The problem statement, all variables and given/known data
An object undergoes simple harmonic motion with its position described by
x(t ) = 50sin (2piet/3)

C) when is Velocity first equal to 80 D) whe nis the Velocity next equal to 80

2. Relevant equations

3. The attempt at a solution

k iknow you have to find the derivative.. which i think is this.

V(t)= 50cos(2.094t)(2.094)
= 104.7cos(2.094t)

then set it to 80: 80=104.7sin(2.094t)
i dont know where to go from here :S

and for D) i have no idea wher to start!

2. Nov 25, 2008

### losthelp

Re: An object goes through simple harmonic motion with its position described as....

if anyone could take a stab at it hatd be great :)

3. Nov 25, 2008

### Staff: Mentor

Re: An object goes through simple harmonic motion with its position described as....

Don't replace 2pi/3 with an approximate value.
You have x(t) = 50 sin((2pi/3)t)
so, v(t) = x'(t) = 50 cos((2pi/3)t) * (2pi/3) = 100 pi/3 * cos((2pi/3)t).

Now, solve for t at the first time when v(t) = 80. Think about how the cosine graph looks when you answer this question and the other one.

4. Nov 25, 2008

### losthelp

Re: An object goes through simple harmonic motion with its position described as....

why does it matter if i make it 100pi/3 * cos((2pi/3)t) ...isnt it the same thing?

well if i solve it like that: 80= 100pi/3 * cos(2pi/3t)
0.7639= cos((2pi/3)t)
2.2918=cos ((2pi)t)
t= 2.918

and then how ould u solve for it next being 80 :S is that just solving the v(t) equation with time at 2.918?

5. Nov 25, 2008

### losthelp

Re: An object goes through simple harmonic motion with its position described as....

mark can you please answer me because im really quite desparate now!

6. Nov 25, 2008

### Staff: Mentor

Re: An object goes through simple harmonic motion with its position described as....

104.7 is not the same number as 100 pi/3. They're not too far apart, but they are different from the 2nd decimal place on.

You will always get more exact results in your answers if you save your approximations for the last step. If you do them too early, you are likely to introduce greater and greater errors, making your final answer even more inaccurate.

80= 100pi/3 * cos(2pi/3t)
0.7639= cos((2pi/3)t)
2.2918=cos ((2pi)t)
t= 2.918

Your step from the 2nd equation to the 3rd is wrong. You apparently multiplied by 3 on both sides, but 3 *cos((2pi/3)t) != cos(2pi t)

Let's start again from the first equation:
100pi/3 * cos(2pi/3t) = 80
cos((2pi/3)t) = 240/(100 pi) = 2.4/pi

Now, what can I do to get (2pi/3)t all by itself on the left side?

If I can do that, then I can multiply both sides by 3/(2pi) to get t.

For t, I get .334890728 for the first solution. Even if you hadn't made a mistake, I think that your answer would have agreed with mine in only three or four decimal places.

For the second solution, you need to know the period of your function and what a cosine function's graph looks like.

7. Nov 25, 2008

### losthelp

Re: An object goes through simple harmonic motion with its position described as....

THANK YOU!!! now the period of my function is 3, and i know how a cosine graph looks like. so could u at all tell me what im suposed to be looking for :S sorry/thanks!

8. Nov 25, 2008

### Staff: Mentor

Re: An object goes through simple harmonic motion with its position described as....

The first time v(t) = 80 at $$t \approx .334890728$$. The next time is when $$t \approx 3 - .334890728$$.

9. Nov 25, 2008

### losthelp

Re: An object goes through simple harmonic motion with its position described as....

thats it? hmmm thank you so much mark! you're a life saver!!!

XOXOXOOXOXOXOXO