# Homework Help: Harmonic Motion; velocity

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1. Aug 11, 2016

### xxphysics

1. The problem statement, all variables and given/known data
The position of a particle undergoing simple harmonic motion is given by x(t)=35cos(10πt), where x is in millimeters and t is in seconds.

Determine the x component of velocity of the particle at t = 0.60 s .

2. Relevant equations
v = x/t

3. The attempt at a solution

I correctly found the x position at time "t" to be 0.035 m (the hw page corrected it to 35 mm). Wouldn't you just take 0.035 and divide by 0.6 to get 0.0583 m/s? I tried that and 58.3 mm/s and they were both wrong. I only have one more attempt at the solution, any help would be appreciated!

2. Aug 11, 2016

### Staff: Mentor

If you are given the position x(t) as a function of time, how do you find the equation for the velocity v(t) as a function of time?

3. Aug 11, 2016

### haruspex

That equation is not correct. The following are correct:
$v_{average}=\Delta x/\Delta t$
$v=\frac{dx}{dt}$

4. Aug 11, 2016

### xxphysics

Intergrate ...?
would the equation be v = A(1/2*t^2)(cos10πt) ?

5. Aug 11, 2016

### Staff: Mentor

No, see haruspex' hint..

6. Aug 11, 2016

### xxphysics

I mean v = A(1/2*t^2)(sin10πt)

7. Aug 11, 2016

### haruspex

No. Not integrate either.

8. Aug 11, 2016

### xxphysics

Derive then? I'm sorry I'm not really sure what the derivative would be?

9. Aug 11, 2016

### haruspex

I do not see how you can solve this problem without being able to differentiate simple trig functions and apply the chain rule. Consult your notes/textbooks.

10. Aug 11, 2016

### xxphysics

I haven't taken calc in three years and my physics book doesn't exactly go over chain rule, that's why I came here for help.
Would you derive with respect to t? v=35*10π*sin(10πt) ?

11. Aug 11, 2016

### Staff: Mentor

Close! Double check what the derivative of cos(t) is...

12. Aug 11, 2016

### xxphysics

Ahh forgot the negative. v=35*10π*-sin(10πt) thank you! For acceleration would you just take the derivative of that for a = 35*10π*10π* -cos(10πt) ? Or since in v=35*10π*-sin(10πt) the sine function equaled zero and the rest didn't really matter was is supposed to be v=10π*-sin(10πt) in which case the derivative would be a=10π*-sin(10πt) ?

13. Aug 11, 2016

### xxphysics

Also, thank you!

14. Aug 11, 2016

### haruspex

Yes.
I do not follow your reasoning. The value of v at some time does not of itself tell you anything about the value of dv/dt at that time. And what "rest"?

15. Aug 11, 2016

### xxphysics

Sorry I thought numbers that didn't have the variable being derived weren't present in the derivation. The reason why you add "+c" to the end of integrations? And since cos(10π*0.6) = 0, whether the rest of the equation was there wouldn't matter since you're multiplying by 0. Anyways thank you :)

16. Aug 11, 2016

### haruspex

Ah, that's what you meant by 'the rest'. Yes, that's right, the only term in the acceleration is the one got by differentiating the sine function wrt time. You do not need to differentiate the coefficients in front of it since t does not feature there.
But you also wrote "since .... the sine function equalled zero" (at time 0, I assume you mean). As I wrote, that has no bearing on the value of its derivative at that time.
Wrong.

17. Aug 11, 2016

### xxphysics

Meant -sin(10π*0.6) = 0, my bad

18. Aug 11, 2016

### haruspex

Ok.
Are you clear on this now? Do you get the right answer?

19. Aug 11, 2016

Yup! Thanks!