Period of Oscillation in a 1D Linear Spring

In summary, the natural oscillation period of a one-dimensional linear spring with spring constant E is 2π.
  • #1
MaxManus
277
1

Homework Statement


The question
I have a one-dimensional linear spring with spring constant E.
The tension is given by σ = Eε, epsilon = strain.. The left side of the spring is held fixed, the right side has a mass m attached to it. We can neglect gravity. What is the natural oscillation period?

What I need help to:
Is it possible to say if it has transverse or longitudinal waves?
Or if I am asking for the wrong hint, please say so.
 
Last edited:
Physics news on Phys.org
  • #2
Hi MaxManus! :smile:
MaxManus said:
Is it possible to say if it has transverse or longitudinal waves?

The question says "linear", so you can assume it's longitudinal. :smile:

Just do an F = ma equation. :wink:
 
  • #3
tiny-tim said:
Hi MaxManus! :smile:The question says "linear", so you can assume it's longitudinal. :smile:

Just do an F = ma equation. :wink:

Thanks, but I didn't get the F=ma hint.

[tex] \sigma = E\epsilon [/tex]
for longitudinal

[tex] \epsilon = \frac{\partial u_x}{\partial x} [/tex]
where u is the displacement field
Longitudinal waves are on the form

u = (ux,0,0)
ux = u0sin(k(x-ct))
where k is the wave number, u0 is the amplitude and c is the phase velocity.
The equation of motion becomes:

[tex] \frac{\partial^2 u_x}{\partial t^2} = \frac{\lambda +\mu}{\rho} \frac{\partial^2 u_x}{\partial x^2} [/tex]
Which gives
[tex] c^2 = \frac{\lambda + 2 \mu}{\rho} [/tex]

And this is as far as I am able to come.
 
Last edited:
  • #4
You are probably confusing yourself by thinking about "waves" at all for this question. It is really about simple harmonic motion of a particle (the mass). Certainly you COULD consider the simple harmonic motion as a standing wave made up of two traveling waves in opposite directions, but that making it a lot more complicated than it needs to be.

Also your use of the word "spring constant" for E seems a bit confused. The equation you give (which is correct) links stress, strain, and Young's Modulus.

What you want is the equation linking force and displacement. That equation involves the length and cross section area of the spring, as well as the value of E.
 
  • #5
Hi MaxManus! :wink:
MaxManus said:
Thanks, but I didn't get the F=ma hint.

You're analysing the whole spring

I meant just look at the mass on the end …

d2x/dt2 = a = F/m …

carry on from there. :smile:
 
  • #6
AlephZero said:
You are probably confusing yourself by thinking about "waves" at all for this question. It is really about simple harmonic motion of a particle (the mass). Certainly you COULD consider the simple harmonic motion as a standing wave made up of two traveling waves in opposite directions, but that making it a lot more complicated than it needs to be.

Also your use of the word "spring constant" for E seems a bit confused. The equation you give (which is correct) links stress, strain, and Young's Modulus.

What you want is the equation linking force and displacement. That equation involves the length and cross section area of the spring, as well as the value of E.

Thanks for the explanation and I now have your equation.
[tex] \sigma = E\epsilon] [/tex]

[tex] \sigma = F\A [/tex]
Wikipedia says:
[tex] \Delta L = \frac{F}{E A} L = \frac{\sigma}{E} L. [/tex]
But not sure what I'm supposed to do with this equation


F=ma
a = d^2 x/\d t^2
 
  • #7
What force does a spring exert when it is compressed by a displacement x? :wink:
 
  • #8
tiny-tim said:
What force does a spring exert when it is compressed by a displacement x? :wink:
Spring force:
F = -kx

d2x/dt2 = a = F/m = -kx/n

x(t) = Acos(sqrt(k/m)*t) + Bsin(sqrt(k/m)*t)

But I haven't used the information I was given
F/A = E*epsilon
 
  • #9
Hi MaxManus! :smile:

(just got up :zzz: …)
MaxManus said:
But I haven't used the information I was given
F/A = E*epsilon

hmm … let's see … the question didn't have a k, instead it had …
MaxManus said:
I have a one-dimensional linear spring with spring constant E.
The tension is given by σ = Eε, epsilon = strain..

tension (σ) is another name for the force, F

strain (ε) = displacement (x) over original length (L)

so F = Eε = (E/L)x, ie k = E/L :smile:
 
  • #10
Thanks again
x(t) = Acos(sqrt(E/(mL))*t) + Bsin(sqrt(E/(mL))*t)

Can I now just do 2pi/T = sqrt(E/(mL))?
T = [tex] 2 \pi \sqrt{\frac{m L}{E}}} [/tex]
 
  • #11
Hi MaxManus! :smile:

(have a pi: π and a square-root: √ :wink:)

let's see … period = time to go 2π …

so it's when √(E/mL)t = 2π, ie t = 2π√(mL/E) …

yup! :biggrin:
 
  • #12
Thank you so much for all the help and patience.
BTW your smileys really cheer me up:smile:
 

FAQ: Period of Oscillation in a 1D Linear Spring

1. What is the period of oscillation in a 1D linear spring?

The period of oscillation in a 1D linear spring is the time it takes for the spring to complete one full cycle of oscillation, from its starting position, through its maximum displacement, and back to its starting position.

2. What factors affect the period of oscillation in a 1D linear spring?

The period of oscillation in a 1D linear spring is affected by the mass of the object attached to the spring, the spring constant, and the amplitude of the oscillation.

3. How is the period of oscillation calculated in a 1D linear spring?

The period of oscillation in a 1D linear spring can be calculated using the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant.

4. How does the period of oscillation change when the mass attached to the spring is increased?

As the mass attached to the spring is increased, the period of oscillation also increases. This is because a larger mass requires more force to be displaced, resulting in a longer period of oscillation.

5. Can the period of oscillation be affected by external factors?

Yes, the period of oscillation in a 1D linear spring can be affected by external factors such as air resistance and friction. These factors can slow down the oscillation and increase the period.

Similar threads

Back
Top