# Acceleration of Mass on Spring

howsockgothap

## 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?

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.

klawlor419
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=w*B=0, B=0
x[t]=A*cos[w*t]
x=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.

klawlor419
I got 15.58 m/s^2 with my calculator.

klawlor419
You used 2.60 in your Cosine, rather than 2.06 ;)

Mentor
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.)

howsockgothap
Thank you! I knew I must have done something ridiculous like that.