What Is the Quality Factor of a Large Foucault Pendulum?

[leonne]

Homework Statement

A large Foucault pendulum such as hangs in many science museums can swing for many hours before it damps out. Taking the decay time to be about 8 hours and length to be 30 meters find the quality factor Q

Homework Equations

q=wo/2b
decaytime=1/b
wo=(k/m)^1/2

The Attempt at a Solution

My friend helped me on this problem and does not seem right
he did t=2pie(l/g)^1/2=10.99 the period of oscillation than found angular velocity w=2pie/t=.57 rad
Then used the formula q=mw/r were r is the damping force and so got the answer as q=.57M/r The formula he used is different from the book so is this wrong or is it just in a different format? or would i just need to change r to 2b= 2(1/8)

 

[mark726]

to get the answer?


Hello! Thank you for posting this question. Let's see if we can figure out the correct solution together.

First, let's clarify the equations that we will be using. The quality factor, Q, is defined as the ratio of the oscillation frequency, ω, to the damping rate, γ. So, the correct equation for Q is:

Q = ω/γ

The decay time, t, is related to the damping rate by the equation:

t = 1/γ

Next, we need to find the natural frequency, ω, of the pendulum. This can be calculated using the equation:

ω = √(k/m)

Where k is the spring constant and m is the mass of the pendulum. However, in this case, we are dealing with a Foucault pendulum, which does not have a spring constant. Instead, we can use the equation:

ω = √(g/L)

Where g is the acceleration due to gravity and L is the length of the pendulum. Plugging in the given values, we get:

ω = √(9.8 m/s^2/30 m) = 0.57 rad/s

Now, we can plug this value into the equation for Q:

Q = ω/γ = 0.57 rad/s / γ

Finally, we can use the given decay time of 8 hours to find the damping rate, γ:

t = 1/γ
8 hours = 1/γ
γ = 1/8 hours^-1 = 0.125 hours^-1

Converting this to seconds^-1, we get:

γ = 0.125 hours^-1 * (1 hour/3600 seconds) = 3.47 x 10^-5 seconds^-1

Now, we can plug this value into the equation for Q:

Q = 0.57 rad/s / 3.47 x 10^-5 seconds^-1 = 1.64 x 10^4

So, the quality factor for this pendulum is approximately 1.64 x 10^4. I hope this helps clarify the correct solution for this problem. Let me know if you have any further questions. Good luck with your studies!