Derive the formula for the frequency of a spring

In summary, the frequency of vibration of the masses along the line connecting them is ω = √[ k(m1 + m2) / (m1*m2)].'
  • #1
astroman707
61
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
  • #2
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:
  • #3
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 Andrew Mason
  • #4
If you use Collinsmark's approach you have to keep in mind that ##x_1## and ##x_2## are the respective displacements of ##m_1## and ##m_2## from the centre of mass of the two-body system.

AM
 

Related to Derive the formula for the frequency of a spring

1. What is the formula for calculating the frequency of a spring?

The formula for the frequency of a spring is given by: f = 1 / (2π√(k/m)), where f is the frequency in hertz (Hz), k is the spring constant in Newtons per meter (N/m), and m is the mass attached to the spring in kilograms (kg).

2. How is the frequency of a spring related to its spring constant?

The frequency of a spring is inversely proportional to its spring constant. This means that as the spring constant increases, the frequency decreases, and vice versa.

3. Can the frequency of a spring be changed?

Yes, the frequency of a spring can be changed by altering either the mass attached to the spring or the spring constant. Increasing the mass will decrease the frequency, while increasing the spring constant will increase the frequency.

4. What are the units of frequency for a spring?

The units of frequency for a spring are hertz (Hz), which is equal to cycles per second. This unit is commonly used for measuring the frequency of vibrations or oscillations.

5. Can the formula for calculating the frequency of a spring be applied to all types of springs?

Yes, the formula for the frequency of a spring can be applied to all types of springs, as long as they exhibit simple harmonic motion. This includes springs with a linear, torsional, or torsion spring constant.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
558
  • Introductory Physics Homework Help
Replies
16
Views
538
  • Introductory Physics Homework Help
Replies
16
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top