Change in period due to a change in mass

[Flipmeister]

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...

 

[haruspex]

You wrote a 'relevant equation' involving T, m and k, but you haven't used it.

 

[nasu]

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?

 

[Flipmeister]

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 $\sqrt{m+Δm}=\sqrt{m} \sqrt{1+\frac{1}{2} \frac{Δm}{m}}$

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...

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

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?

 

[Flipmeister]

Just realized this problem had a solution for it in the book, and that answer is right. Thanks for the help!

 

[nasu]

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.