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SHM Question - starting at x=A, find when passes through x=0

  • Thread starter Glorzifen
  • Start date
  • #1
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Homework Statement


A mass oscillating in SHM starts at x=A and has a period of T. At what time, as a fraction of T, does it first pass through x=0?

Homework Equations


x(t) = Acos(wt + [tex]\varphi[/tex])

The Attempt at a Solution


I'm not sure if this means I set x(t) = A, or if the [tex]\varphi[/tex] = -A?
Also, does it use A because I'm supposed to recognize that it's equal to amplitude and do something about it (i.e. cancel)? Not too sure how to go about doing this though...any hints?
 

Answers and Replies

  • #2
53
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The period is the amount of time it takes for a SHM to complete one cycle. Draw out x(t) and look at what fraction of a cycle it takes to reach first reach x=0. Then from this think about the fraction of a period it took for this to happen.

Don't get confused by that phase shift variable. It probably wont be important in your introductory physics class. anyways the problem tells us at t=0, x=A or x(0) = A. for this to be true the phase shift must be zero as well. (Cos(0) =1)
 
  • #3
25
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The period is the amount of time it takes for a SHM to complete one cycle. Draw out x(t) and look at what fraction of a cycle it takes to reach first reach x=0. Then from this think about the fraction of a period it took for this to happen.
Oh...is it saying that x=A as in I'm to 'start' with wherever there is a max amplitude? In which case the answer is 1/4T. I'm not sure if the A in x=A is the same A is amplitude...I'm not sure what else it would mean though.
 
  • #4
53
0
Yes that's right.
But the A is the same through all of the problem. in the formula for something showing SHM, x(t) = A Cos(wt - phi), The A stands for the maximum amplitude of the object. by saying x=A we are just saying that cos has reached is maximum value (1). which if you visualize pulling a spring or something and letting go, when we start our time (t=0) the amplitude is the largest it will ever be, or x=A.
 

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