# Determining g from harmonic frequency and a pendulum

1. Jan 31, 2008

### vertabatt

1. The problem statement, all variables and given/known data
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.83g/m and a 1.21 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.21 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200Hz . Next, with the 1.21 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 100 oscillations. Pulling out your trusty calculator, you get to work.

What value of g will you report back to headquarters?

2. Relevant equations
F(1) = 1/2L * √(Tension/linear density)
Tension = 2*$$\pi$$ * √(L/g)

3. The attempt at a solution
First I drew a picture of the vibrating wire and labeled each half of the wire d.

So, since F(2) = 2F(1):

F(1)= 100Hz = 1/2d * √(1.21g/.00183)

Using this I solved for g:

g = 60.50d^2

Then I moved on to the pendulum portion, calling 2d the length of the wire that the mass hangs from:

Period = oscillations/time = 100/313s = .3195

Period = 2*$$\pi$$√(L/g), so:

.3195 = 2*$$\pi$$√(2d/g)

Solving for d:

d = .001293g

Then I combined equations to solve for g:

60.50(1.0111*10^-4g)^2 = g

g(1.0111*10^-4 - 1) = 0

g = 0, 9889.97

0 is an unreasonable answer, so I am left with 9889.97m/s^2, which is also pretty unreasonable (and incorrect).

What am I doing wrong here?

Last edited: Jan 31, 2008
2. Jan 31, 2008

### vertabatt

I've repeated this result a few times... leading me to believe that possibly my pendulum calculation is incorrect because the mass of the wire has not been taken into account.

The length L for the pendulum is the length to the center of mass, perhaps that length isn't 2d? Is there another equation that might help me solve this?

3. Jan 31, 2008

### Werg22

Do you know anything about the length of the pendulum? If so, why not calculate the position of the center of mass of the wire + 1.21 kg weight system and treat the problem as a simple, with massless wire, pendulum problem?

Last edited: Jan 31, 2008
4. Feb 1, 2008

### vertabatt

Nothing was given about the pendulum length. I actually did the problem correctly except that I used 100/313 to find the period when I should have used 313/100.

The correct answer was g = 1.07m/s^2