Find Acceleration At Time t For Mass on a Spring

  • Thread starter Becca93
  • Start date
  • #1
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Homework Statement
A mass is oscillating on a spring with a period of 4.35 s. At t = 0 the mass has zero speed and is at x = 3.25 cm. What is the magnitude of the acceleration at t = 1.30 s?

The attempt at a solution

The formula for oscillatory acceleration is
a(t) = -Aω^2cos(ωt - ∅)
a(x) = -(ω^2)x

ω = 2pi/period, which is 2pi/4.35

So,

a(t) = -(2pi/4.35)(3025cos((2pi/4.35)(1.30)))
a(t) = 0.6798 m/s^2

This isn't the correct answer.

The question states that it starts at t= 0 at x=3.25, so I took that to be A. Is that my problem? Where am I going wrong?
 

Answers and Replies

  • #2
1,198
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Check your arithmetic. I assume 3025 is actually 3.25. Secondly, you are not evaluating the formula you cite.
 
  • #3
Doc Al
Mentor
45,248
1,594
ω = 2pi/period, which is 2pi/4.35
OK.
So,

a(t) = -(2pi/4.35)(3025cos((2pi/4.35)(1.30)))
Redo this.
The question states that it starts at t= 0 at x=3.25, so I took that to be A. Is that my problem?
That makes sense to me. (Be sure to use standard units for distance.)
 
  • #4
84
1
I have the answer now. Thank you!
 

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