Change in period due to a change in mass

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Homework Help Overview

The discussion revolves around the change in the period of a mass oscillating on a spring when the mass is slightly increased. The original poster seeks to express the change in period, ΔT, in terms of the initial period T, the mass m, and the change in mass Δm, while avoiding the spring constant k.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive an expression for ΔT starting from the equation for the period T. They express uncertainty about how to eliminate the spring constant k from their calculations. Some participants suggest using the binomial approximation for small changes in mass and question the validity of certain assumptions made during the derivation.

Discussion Status

Participants have engaged in a back-and-forth regarding the derivation of ΔT, with some providing guidance on using approximations. There is acknowledgment of the original poster's approach as a good start, and they are encouraged to explore further simplifications. The discussion reflects a collaborative effort to clarify the reasoning behind the equations involved.

Contextual Notes

There is a focus on the condition that Δm is much smaller than m, which is critical for the approximations being discussed. Additionally, some participants note that certain assumptions made during the discussion may not be justified by the problem's requirements.

Flipmeister
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Homework Statement


A mass m oscillating on a spring has a period T. Suppose the mass changes very slightly from m to m+Δm, where Δm << m. Find an expression for ΔT, the small change in the period. Your expression should involve T, m, and Δm, but NOT the spring constant (k).


Homework Equations


T=\frac{2\pi }{\omega}=2\pi \sqrt{\frac{m}{k}}\\<br /> \omega=\sqrt{\frac{k}{m}}


The Attempt at a Solution


T+ΔT=2\pi \sqrt{\frac{m+Δm}{k}}
Would that be the right way to start off? I can't figure out how to get rid of the k there. I don't see how conservation of energy can help, especially considering that conservation of energy is how those equations were derived in the first place...
 
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You wrote a 'relevant equation' involving T, m and k, but you haven't used it.
 
Flipmeister said:

The Attempt at a Solution


T+ΔT=2\pi \sqrt{\frac{m+Δm}{k}}
Would that be the right way to start off? I can't figure out how to get rid of the k there. I don't see how conservation of energy can help, especially considering that conservation of energy is how those equations were derived in the first place...
Yes, this is a good start.
You need to use now the fact that Δm<<m and find an approximate expression of the RHS.
Do you know the expansion of \sqrt{1+x} for small values of x?
 
nasu said:
Yes, this is a good start.
You need to use now the fact that Δm<<m and find an approximate expression of the RHS.
Do you know the expansion of \sqrt{1+x} for small values of x?

Yes, using the binomial approximation, I get ##\sqrt{m+Δm}=\sqrt{m} \sqrt{1+\frac{1}{2} \frac{Δm}{m}}##

haruspex said:
You wrote a 'relevant equation' involving T, m and k, but you haven't used it.
Hmm I suppose I should find T and ΔT separately next and plug those in...

T+ΔT=2π(1+\frac{Δm}{2m})(\sqrt{m/k})=2π\sqrt{m/k} + 2π\sqrt{Δm/k}<br />

Simplifying I get √(m/k) = 2, so T=4π.

Solving for ΔT, I get ##ΔT=2π(1+(1/2) \frac{\Delta m}{m}) \sqrt{m/k} - T## which gets me...
$$\Delta T = 4π + 2π \frac{\Delta m}{m} -4π= T(\frac{1}{2} \frac{\Delta m}{m})$$

Does this look right? It seems to make sense (increasing the mass should increase period according to the starting equations). Is there any way for me to confirm this?
 
Last edited:
Just realized this problem had a solution for it in the book, and that answer is right. Thanks for the help!
 
Flipmeister said:
Simplifying I get √(m/k) = 2, so T=4π.

This step is not justified by the problem and you don't need this assumption to solve the problem. 2π√(m/k) = T, where T is the initial period.
However, the final result is OK.
 

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