What is the Effect of Distributed Mass on the Time Period of a Spring?

In summary, the conversation discusses the time period of a vertically hung spring with mass 'M' and spring constant 'K', divided into n equal massless springs connected by n small objects. The extension in the first spring is determined to be Δx1=mg/k and the total extension is found to be ΔX=Mg n(n+1)/2k. However, this does not apply to the dynamic case of oscillations, where the effective mass is actually M/2 rather than M/3 as commonly stated. The difference is due to the spring's oscillations being in the form of standing waves, making the mathematics more complex.
  • #1
ShakyAsh
6
0
Okay, so I've been trying to figure out what would be the time period of spring of mass 'M' & spring constant K which is hung vertically from a ceiling.

Dividing the spring into n equal massless springs of spring constant k=nK which are connected to each other by n small objects of mass m=M/n like this-
_____
n | (k)
m
:
:
3 | (k)
m
2 | (k)
m
1 | (k)
m​
Now, extension in first spring
Δx1=mg/k
Similarly,
Δx2=2mg/k
...
Δxn=n mg/k
ΔX(Total)=ƩΔx=mg n(n+1)/2k
Putting values of m and k
ΔX=Mg n(n+1)/2Kn^2

Taking the limit n→∞
ΔX=Mg/2K which corresponds to an effective mass of M/2 rather than M/3 as given in Wikipedia & other websites.

I would be grateful if someone would be able to explain what i am doing wrong.
 
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  • #2
I think what you are doing wrong is to assume that the effective mass of the spring for static extension (equilibrium) is also the effective mass for the dynamic case (oscillations).

The oscillations of a spring with distributed mass will be in the form of standing (stationary) waves. The case you are probably interested in is the fundamental mode. The mathematics is more complicated than for a lumped mass on a massless spring.
 

1. What is the effective mass of a spring?

The effective mass of a spring refers to the mass that needs to be attached to the spring in order for it to behave like a simple harmonic oscillator.

2. How is the effective mass of a spring calculated?

The effective mass of a spring can be calculated by dividing the spring constant (k) by the square of the angular frequency (ω) of the oscillations.

3. What factors affect the effective mass of a spring?

The effective mass of a spring is affected by the mass of the spring itself, as well as any additional masses attached to it. It is also influenced by the stiffness of the spring and the frequency of the oscillations.

4. Why is the effective mass of a spring important?

The effective mass of a spring is important because it allows us to simplify the analysis of a complex system by treating it as a simple harmonic oscillator. This makes it easier to understand and predict the behavior of the system.

5. How does the effective mass of a spring relate to its natural frequency?

The effective mass of a spring is directly proportional to its natural frequency. This means that as the effective mass increases, the natural frequency of the spring also increases, and vice versa.

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