Determining g from harmonic frequency and a pendulum

• vertabatt
In summary, as the captain of a scientific team on Planet Physics, you were tasked with measuring g. With a long, thin wire labeled 1.83g/m and a 1.21kg weight, you were able to find the midpoint of the wire by folding it in half and attach it to the wall of your laboratory. After attaching the weight to the end of the wire and vibrating it, you found the wire's second harmonic frequency to be 200Hz. Then, by making a pendulum with the wire and measuring the time for 100 oscillations, you were able to calculate the value of g to be 1.07m/s^2. You also realized that the length of the pendulum was not given
vertabatt

Homework Statement

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?

Homework Equations

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

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:
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?

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:
Werg22 said:
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?

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

1. What is the equation for calculating the acceleration due to gravity using a pendulum?

The equation for calculating the acceleration due to gravity using a pendulum is g = (4π^2L) / t^2, where g is the acceleration due to gravity, L is the length of the pendulum, and t is the period of the pendulum's oscillation.

2. How is the length of the pendulum related to the acceleration due to gravity?

According to the equation mentioned above, the length of the pendulum is directly proportional to the acceleration due to gravity. This means that as the length of the pendulum increases, the acceleration due to gravity also increases, and vice versa.

3. What is the period of a pendulum?

The period of a pendulum is the time it takes for one full oscillation, or swing, from one side to the other. It is typically measured in seconds.

4. How can the frequency of a pendulum be used to determine the acceleration due to gravity?

The frequency of a pendulum is the number of oscillations it makes per unit of time. By using the equation g = (4π^2L) / t^2, we can rearrange for t and plug in the frequency to solve for the acceleration due to gravity.

5. What factors can affect the accuracy of calculating g using a pendulum?

Some factors that can affect the accuracy of calculating g using a pendulum include air resistance, friction, and the precision of the measurement of the pendulum's length and period. Additionally, the location and altitude at which the experiment is conducted can also impact the measurement of g.

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