Change in period due to a change in mass

AI Thread Summary
A mass m oscillating on a spring has a period T, which changes slightly when the mass is increased to m + Δm. The discussion focuses on deriving an expression for the small change in period, ΔT, without involving the spring constant k. Using the binomial approximation for small changes, the relationship between ΔT and the original period T is established as ΔT = T(1/2)(Δm/m). The final expression confirms that an increase in mass leads to an increase in the oscillation period, aligning with the expected physical behavior. The solution is validated against a textbook answer, confirming its correctness.
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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