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Simple Harmonic Motion-Finding the distance traveled by the vibrating object?

  • #1
Simple Harmonic Motion-Finding the distance traveled by the vibrating object???

Hi, I would really like to know if my solution to the following question is correct, I would really really really appreciate it.

A hanging spring stretches by 35cm when an object of mass 450g is hung on it at rest. In this situation, we define its position as x=0. The object is pulled down an additional 18cm and released from rest to oscillate without friction. What is its position x at a time 84.4s later?
Find the distance traveled by the vibrating object.

Well, by using the equation -ky-mg=0 (at rest) I derived the equation
y=-mg/k. So i replaced y with 35 cm and m with 0.45kg in order to find k.
With k, i set the equation: x(t)= Acos(wt+c) c=phase constant w=angular frequency
and since this started at rest, there is no phase constant, and amplitude is 18cm so i made the equation
x(t)=0.18cos(5.29t)
then i replace the t with 84.4. Is this right?
And how do i go about finding the "distance traveled by the vibrating object"?
Thanks!
 
Last edited:

Answers and Replies

  • #2
574
0
Just calculate how many oscillations it goes through and multiply the amplitude with that.
 
  • #3
Fermat
Homework Helper
872
1
That should be the double-amplitude.
 
  • #4
andrevdh
Homework Helper
2,128
116
I think you got it right if we agree that positive x is measured downwards.
For the distance travelled you can try T = 2π/ω for the period of the oscillation. Dividing the time by the period wil therefore give the amount of full oscillations (I get 71) plus a bit (0.0790 of an oscillation which converts to a time of 0.0938s). For every full oscillation the mass covers four times the amplitude. Using your equation and the amount of leftover time you can calculate the additional disance it travelled.
 

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