Finding Velocity in Simple Harmonic Motion

[xxphysics]

Homework Statement

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 .

Homework Equations

v = x/t

The Attempt at a Solution

I correctly found the x position at time "t" to be 0.035 m (the homework 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!

 

[berkeman]

Homework Statement

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 .

Homework Equations

v = x/t

The Attempt at a Solution

I correctly found the x position at time "t" to be 0.035 m (the homework 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!

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?

 

[haruspex]

v = x/t

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

 

[xxphysics]

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?

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

 

[berkeman]

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

No, see haruspex' hint..

 

[xxphysics]

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

I mean v = A(1/2*t^2)(sin10πt)

 

[haruspex]

Intergrate

No. Not integrate either.

 

[xxphysics]

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

 

[haruspex]

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

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.

 

[xxphysics]

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.

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) ?

 

[berkeman]

x(t)=35cos(10πt)

v=35*10π*sin(10πt) ?

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

 

[xxphysics]

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

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) ?

 

[xxphysics]

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

Also, thank you!

 

[haruspex]

a = 35*10π*10π* -cos(10πt) ?

Yes.

Or since in v=35*10π*-sin(10πt) the sine function equaled zero and the rest didn't really matter

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"?

 

[xxphysics]

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"?

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 :)

 

[haruspex]

Sorry I thought numbers that didn't have the variable being derived weren't present in the derivation

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.

cos(10π*0.6) = 0

Wrong.

 

[xxphysics]

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.

Meant -sin(10π*0.6) = 0, my bad

 

[haruspex]

Meant -sin(10π*0.6) = 0, my bad

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

 

[xxphysics]

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

Yup! Thanks!