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Springs: speed and acceleration

  • Thread starter Sheneron
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1. Homework Statement
A 0.450 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 11.0 cm.
Calculate the maximum value (magnitude) of its speed and acceleration.[/B]

3. The Attempt at a Solution

I found both w and t.

[tex] \omega = 4.216[/tex]
[tex] t = 0.145[/tex]

To get t I solved this equation for t.

[tex] x(t) = Acos(\omega t + \phi)[/tex]
[tex]t = \frac{1}{4.216}cos^-1(\frac{x}{11})[/tex]


With that value of t, I solved for v. I got a negative number which is incorrect. However, if at the start I use sine instead of cosine for position i get a different value for t, and get the same answer for v, except positive, which is correct. But why would I use sine for position rather then cosine, and how would I know? Also why would it be positive or negative?
 

Answers and Replies

Dick
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Why are you solving for 't'. What t? Just put x(t)=A*cos(omega*t). Then v(t)=x'(t) and a(t)=x''(t). Work those out. Now to the find the maxes you don't even have to worry about the trig function. The max of either sin or cos is 1.
 
360
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I am sorry... I left off part of the question...

(b) Calculate the speed and acceleration when the object is 9.00 cm from the equilibrium position.

That is why I was finding t
 
Dick
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I am sorry... I left off part of the question...

(b) Calculate the speed and acceleration when the object is 9.00 cm from the equilibrium position.

That is why I was finding t
Ah. In that case the problem doesn't give you enough information to determine a sign for v or a. The object could be 9 cm from equilibrium and either moving away from or towards equilibrium. They seem to just want the magnitude of v and a. Not the sign. You can use either sin or cos. They only differ by a change the phase angle (phi).
 
alphysicist
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As an alternative, you still are not required to find the time for part b. You can use conservation of energy to find the speed and a force diagram for the magnitude of the acceleration.
 
360
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Ah. In that case the problem doesn't give you enough information to determine a sign for v or a. The object could be 9 cm from equilibrium and either moving away from or towards equilibrium. They seem to just want the magnitude of v and a. Not the sign. You can use either sin or cos. They only differ by a change the phase angle (phi).
Yes thats what thought, but I submitted Acceleration with a postive sign, and it was counted wrong, but with a negative sign correct.

As an alternative, you still are not required to find the time for part b. You can use conservation of energy to find the speed and a force diagram for the magnitude of the acceleration.
I'll try this method thank you.
 
360
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Also, the thing I didn't understand was... I have a solutions manual and this problem was in there. I checked the book and it started with x(t) = Asin(wt + phi), rather then cosine. Which gave the correct sign for direction. But how would you know whether to use cosine or sine? Generally you use cosine don't you?
 
Dick
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That's silly. If they had wanted you to get that specific setup, they should have told you. sin(x+pi/2)=cos(x). There is no difference except for the phase. Even if you use sin, there are 4 times every cycle when the object is 9cm away from equilibrium - and they have all possible combinations of signs for a and v.
 
360
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Well, Thanks for your help.
 

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