Derive the formula for the frequency of a spring

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
astroman707
Messages
61
Reaction score
5

Homework Statement


Two masses m1 and m2 are joined by a spring of spring constant k. Show that the frequency of vibration of these masses along the line connecting them is
ω = √[ k(m1 + m2) / (m1*m2) ]
(Hint: Center of mass remains at rest.)

Homework Equations


f = w/2π
w = √(k/m)
F = -kx
a = - w2x

The Attempt at a Solution


I tried finding the center of mass and using that as m, in ma = kx
I also tried manipulating the formula for acceleration, and plugging it into hooke's law, but that didn"t seem right either. I'm pretty lost to be honest.
 
Physics news on Phys.org
I take it that you know that for a fixed spring on one end and a mass m on the other that the frequency of oscillation is ##\omega = \sqrt{\frac{k}{m}}##

If you try to work it out as if one of the masses was fixed and another notional mass vibrating relative to it such that the acceleration of that mass is equal to the relative acceleration between the two masses m1 and m2, you can solve the problem. Call the notional single mass that is vibrating ##\mu##. If you set ##\mu a_{rel} = -kx## where ##a_{rel} = a_2 - a_1##, you can see that ##\omega = \sqrt{\frac{k}{\mu}}##. You just have to determine what ##\mu## is. (hint: by Newton's third law ##m_1a_1 = - m_2a_2##)

AM
 
Last edited:
Hello @astroman707,

'Just curious, is this problem from coursework that requires calculus or differential equations?

Anyway, there's another way to approach this problem if you don't wish to use relative accelerations.

The first order of business is to define your displacements and then relate them to the stretch of the spring. For example, if you choose to define that [itex]x_1[/itex] is positive when [itex]m_1[/itex] moves to the left and [itex]x_2[/itex] is positive when [itex]m_2[/itex] moves to the right, then [itex]x = x_1 + x_2[/itex]. On the other hand, if you want to define positive in the same x-direction for both, then there will be a negative sign in your equation somewhere (e.g., [itex]x = x_2 - x_1[/itex]). Anyway, the choice is yours, but you'll need to define your terms before we move on.

With that, you should have enough to form two differential equations, one for each mass, by using Newton's second law. But don't worry, you'll only need to use one of them.

The trick then is to use the hint that @Andrew Mason made in the previous post. You're interim goal is to find a relationship of the ratio [itex]\frac{x_1}{x_2}[/itex] in terms of [itex]m_1[/itex] and [itex]m_2[/itex]. Andrew's hint about Newton's third law will get you there.

After that, do a little substitution and you'll have all you need to solve either of the second order, ordinary differential equations and you'll have your [itex]\omega[/itex].
 
  • Like
Likes   Reactions: Andrew Mason