Find Acceleration At Time t For Mass on a Spring

[Becca93]

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?

 

[LawrenceC]

Check your arithmetic. I assume 3025 is actually 3.25. Secondly, you are not evaluating the formula you cite.

 

[Doc Al]

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.)

 

[Becca93]

I have the answer now. Thank you!

 

[asteriatic]

I would first check to see if the given values are consistent with the formula being used. In this case, the period of 4.35 s and the amplitude of 3.25 cm are not consistent with the formula for oscillatory acceleration. The formula you have used is for simple harmonic motion, where the amplitude is equal to the maximum displacement of the mass from its equilibrium position. However, in this problem, the amplitude is given as 3.25 cm but the maximum displacement is not mentioned, so it is not clear if the mass is starting from its equilibrium position or not.

Assuming that the mass is starting from its equilibrium position at t=0, the correct formula to use would be a(t) = -(ω^2)x, where x is the displacement at time t. In this case, the displacement at t=1.30 s would be 3.25 cm, so the acceleration would be:

a(t) = -(2pi/4.35)^2(3.25 cm) = -5.91 m/s^2

This is a negative value because the mass is moving towards its equilibrium position at t=1.30 s.

In conclusion, the incorrect value obtained in the attempt at a solution was due to using an incorrect formula for the given scenario. It is important to carefully check the given values and make sure they are consistent with the formula being used.