Homework Help: Odd equation relating to springs and SHM

1. Apr 26, 2005

K - Prime

Well im working on a lab report for 12th grade physics and my teacher has decided to throw a cuve ball at us. The lab is designed to determine the value of the force constant of a spring by measuring the number of oscillations a certain weight makes when on the end of the spring. To find k was easy, i just used the equation T = 2pi (m/k)^1/2. Heres the part i dont get...he asked us to find the value of m/3 using a graph of T^2 vs m (mass on spring) and gave us the equation T = 2pi [(M + m/3)/k]^1/2. All i can say is WHAT!?

2. Apr 27, 2005

whozum

Well the point ofthe lab wasto find the spring constant, and you modeled that already by graphing it, correct? (If not, do so). Now since you know the trend of period vs mass, you should be able to predict mass ifyou measure a certain period. Notice that the only difference between the firstequation you gave and the last one (which you didnt recognize) is that the last one has a different value for mass.

$$T = 2\pi\sqrt{\frac{m}{k}}$$

He gives:

$$T = 2\pi\sqrt{\frac{M+m/3}{k}}$$

If you say the mass is $n$, then $n = M + m/3$ and the second equation becmoes

$$T = 2\pi\sqrt{\frac{n}{k}}$$, which is analogous to the first equation. So the numerator under the square root is simply a new total mass, nothing to be surprised about. Using the plot you derived in your original lab, find a way to calculate the new mass.

This process isnt necessary to do the problem, but hopefully it helps you see what is going on more clearly.

3. Apr 27, 2005

K - Prime

ahh i see, that helps alot! thx!