Finding the acceleration-displacement equation from potential energy

[thunderhadron]

Hi friends the problem is -

https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/30370_2656498989013_1471109032_n.jpg

Attempt -

friends as per the question I am trying to get the acceleration- displacement equation for this problem. So I am using

F = - (dU / dx)
Differentiating Potential energy function w.r.t. x I get,

F = - (2a + 4b .x³)

But F = mass. acceleration so,

acceleration = - (2a/m + 4b/m x³)

Now I am sticking here that how to proceed further to get the result like,

acceleration = - ω² . x

Please friends help me in solving this Problem.

Thank you all in advance.

 

[TSny]

Check to make sure that you took the derivative of ax² correctly. Also keep in mind that you are considering "small" oscillations.

 

[thunderhadron]

Check to make sure that you took the derivative of ax² correctly. Also keep in mind that you are considering "small" oscillations.

F = - (dU / dx)
Differentiating Potential energy function w.r.t. x I get,

F = - (2ax + 4b .x³)

But F = mass. acceleration so,

acceleration = - (2a/m.x + 4b/m x³)

It is for small oscillations so x³ will be neglected.

Hence accⁿ = -(2a/m).x

acceleration = - ω² . x

hence The answer comes, ω = √(2a/m)

Yet the answer is not achieved. In the book the answer is option (B)

 

[TSny]

I think you got the right answer. I don't see how (B) could be the answer.

 

[thunderhadron]

I think you got the right answer. I don't see how (B) could be the answer.

Yes there is mistake in the answer of book.