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Springs and simple harmonic motion

  • Thread starter Jm4872
  • Start date
  • #1
12
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Homework Statement



A mass is oscillating on a spring with a period of 1.45 s. At t = 0 the mass has zero speed and is at x = 8.80 cm. What is the magnitude of the acceleration at t = 2.30 s?

Homework Equations


x(t)=Acos(wt + phi)
a(t)=-Aw^2cos(wt + phi)
a=w^2x
T=2pi/w


The Attempt at a Solution



I've tried this a few different ways, first off what I do is convert the period (1.45s) to angular velocity using T=2pi/w, this then gives me the result of 4.33 rad/s. Using this I then substitute values into the equation for position to get,
8.80 = Acos(0)
from this equation I then solve for A and recieve 8.80 cm.
Now the information I have is..
w=4.33 rad/s
A=8.80cm
t=2.30s
All of this I put into my equation for acceleration to get..
a(2.30)=-8.80(4.33^2)cos(4.33*2.30)
solving for acceleration the result comes to be 142 m/s^2, however when I input this number it comes out to be wrong? I'm confused as to what it is that I'm doing wrong, please help!
 

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
36
You just forgot to convert the amplitude of the oscillation (which is in cm) to metres before computing your answer.
 
  • #3
12
0
oh wow haha now I feel dumb!
haha thank you!
 

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