Acceleration of Mass on Spring

  • #1
howsockgothap
59
0

Homework Statement


A small block is oscillating on a spring with a period of 3.05 s. At t = 0 the mass has zero speed and is at x = 4.35 cm. What is the magnitude of the acceleration at t = 4.30 s?


Homework Equations


a=-Aω2cos(ωt)
ω=2∏/T

The Attempt at a Solution


ω=2∏/3.05=2.06
x(0)=Acos(0)=4.35=A(1)=A
so:a=(t)=-4.35(2.06)2cos(2.60(4.30))=-18.3 m/s2

which I've been told is wrong.

I have no idea what I'm doing wrong, I'm sure it's something painfully obvious but I just can't figure it out.
 

Answers and Replies

  • #2
klawlor419
117
0
F=-k x[t]
x''[t]=-(k/m)x[t]
=-(w^2)x[t]
x[t]=A*cos[w*t]+B*Sin[w*t]
v[t]=w*(-A*sin[w*t]+B*cos[w*t])
a[t]=-w^2*(A*cos[w*t]+B*Sin[w*t])

v[0]=w*B=0, B=0
x[t]=A*cos[w*t]
x[0]=A=4.35

a[t]=-w^2*(4.35*cos[w*t])
w=2Pi/T

As long as you did the calculator work correctly you should have the right answer.
 
  • #3
klawlor419
117
0
I got 15.58 m/s^2 with my calculator.
 
  • #4
klawlor419
117
0
You used 2.60 in your Cosine, rather than 2.06 ;)
 
  • #5
Doc Al
Mentor
45,412
1,849
so:a=(t)=-4.35(2.06)2cos(2.60(4.30))=-18.3 m/s2
Note that the angle within the cosine is in radians. (That might be the issue if your calculator was set to degree mode.) And it should be 2.06, not 2.6. (Perhaps that was just a typo.)
 
  • #6
howsockgothap
59
0
Thank you! I knew I must have done something ridiculous like that.
 

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