# Q Factor & Run Time of Pendulum

1. Oct 24, 2013

### oddjobmj

1. The problem statement, all variables and given/known data
Consider a small cuckoo clock in which the length of the pendulum is L, suspending a mass m with a period of oscillation T. The clock is powered by a falling weight (mass = M) which falls 2 m between the daily windings. The amplitude of the swing is 0.2 radians.

(A) Determine the quality factor Q of the clock.

Data: m = 55 g; T = 1.0 s; M = 1066 g.

(B) How many days would the clock run if it were powered by a 1.5 volt battery with charge capacity equal to 1800 mAh?

2. Relevant equations

Q=2*$\pi$*$\frac{stored}{lost/cycle}$

Time=$\frac{(mAh/1000)*(voltage)}{device power}$

Power=J/s

3. The attempt at a solution

For part A my understanding is that the energy stored is the potential energy of M:

m*g*h=(1.066 kg)*9.81 (m/s2)*(2 meters)=20.915 Joules

Also, because that energy is dissipated each day, the energy lost per cycle (which is 1 second) is the stored energy (20.915 Joules) divided by the number of seconds in a day (24*60*60) or:

2.42*10-4 Joules

If I plug these values into my equation for Q I get a huge, nonsense number. What am I doing wrong?

For part B:
I have mAh and volts but I need power and what I'm getting for J/s above must be wrong.

Thank you!

Last edited: Oct 24, 2013
2. Oct 24, 2013

### Staff: Mentor

The energy stored per cycle of the clock is related to the swing of the pendulum. I'd look at the potential energy of the pendulum bob when its KE is zero...

Your energy per cycle of 2.42 x 10-4 J represents the energy lost each cycle that's replenished by the falling weight.

For part B, with the amp-hour and voltage rating of the battery you have the total energy stored in the battery. How many day's weight-falls does that energy represent?

3. Oct 24, 2013

### oddjobmj

Ahh, thank you!

Well, the potential when K=0 is dictated by the length of the pendulum and the bob's mass.

T=2π$\sqrt{L/g}$

So, L=0.24849 meters

Using that in the eq for U:

U=mgL(1-cos(.2))=.002673

Plugging that in as the numerator in my eq for Q:

Q=69.4 which is correct!

For part B:

Energy=V*Amps*3600 seconds
Time=Energy/(2*1.066*9.81)=465 days

Thanks again!

4. Oct 24, 2013