# What is the correct formula for acceleration for SHM

1. Feb 20, 2015

### dilton_8000

1. The problem statement, all variables and given/known data
If the mass of a horizontal mass-spring oscillator undergoing SHM is 0.5kg and the force constant is 20N/cm, what is the maximum restoring force of the oscillator? And what is the maximum acceleration? (Time period is 6s and amplitude of oscillation is 12cm)

2. Relevant equations
Acceleration (a) = -w^2(x) and a = -k(x/m)

3. The attempt at a solution
When I use the given information and plug in the values, the answer is different for each equation. For the first equation, I get an answer of 0.13ms^-2, and for the latter I get 480ms^-2.

I don't seem to understand why they don't match and which one is correct. because I know we had to equate ma=-kx to derive the equation omega(w) = sqr.rt(k/m).

Any replies would be tremendously helpful. Thanks.

PS - This is my first post. :)

2. Feb 20, 2015

### Nathanael

Welcome to PF

I'm not entirely sure I understand what's confusing you:
You calculated the acceleration using a=-ω2x and got a different answer than when you used a=-kx/m?

But like you said, ω=√(k/m) so how can the equations be any different?

3. Feb 20, 2015

### dilton_8000

Thanks for the reply.... Actually Yes, I did get two different answers for the different formulae. I think the problem lies with the value to ω. When I use

ω=√(k/m),
with k = 20N/cm, and m = 0.5kg,

Plugging in this value of ω in

a=-ω^2x gives me 4.80 ms^-2

But when I use

ω = 2π/T, with T = 6s,
I get the value 1.05 rad s^-1.

Thus, a=-ω^2x now gives me 0.13 ms^-2

I can't figure out why I get two different values. Shouldn't they match? And also, which one is correct?

PS - I used F =-k.x to find the maximum restoring force. I got 240 N which seems to be correct. Its the acceleration bit thats throwing me off.

Last edited: Feb 20, 2015
4. Feb 20, 2015

### Staff: Mentor

The given spring constant and mass imply a natural period of about a tenth of a second. This is a far cry from the suggested 6 seconds. No mention of damping is made (and it would have to be large indeed), so the given values seem self-contradictory.

5. Feb 21, 2015

### dilton_8000

Thanks so much..... I was suspecting something must have been wrong with the question. BTW, when you say the information implies a natural period of a tenth of a second, how do you arrive at that conclusion? I don't think I know of any equations or relationships that helps me calculate that. Can you help? :)

6. Feb 21, 2015

### Staff: Mentor

Angular frequency ω is related to the period. You should know how ω relates to f, and how f relates to T.

7. Feb 21, 2015

### dilton_8000

Oh, ofcourse, silly me... :)... Thanks... I think I've figured it out... Cheers