Grandfather's Clock - Nonlinear Pendulum

[m0nk3y]

Homework Statement

A Grandfather's Clock is adjusted to keep perfect time if the amplitude is 2 degrees. How many seconds does it lose or gain in a year if instead the amplitude is 10 degrees.

Homework Equations

T/4 = integral from 0 to qmax = dq/$$\sqrt{2E-q^2 - \epsilon/2*q^4}$$
where q is position
E= 1/2 (q'^2 + q^2) + $$\epsilon/4 * q^4$$

The Attempt at a Solution

I started by converting 2 degrees and 10 degrees into radians and, then I took the difference I got 2*pi/45 , and made the integral from 0 to 2*pi/45 since we want to get the time difference between these angles, but I don't know where to go from here, and I don't know if what I did was right.

Thanks

 

[Jhero]

for your question! It looks like you're on the right track with converting the angles to radians and setting up an integral. However, the integral you provided is for the period of a simple harmonic oscillator, which may not be applicable to this situation.

To calculate the time difference between 2 degrees and 10 degrees, we can use the equation for the period of a simple pendulum: T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. In this case, the length of the pendulum would be the same, but the amplitude would change from 2 degrees to 10 degrees.

So, the new period would be T' = 2π√(L/g') where g' is the acceleration due to gravity at an amplitude of 10 degrees. To find g', we can use the equation g' = g(1 + (sinθ)^2), where θ is the amplitude in radians.

Plugging in the values, we get g' = 9.8(1 + (sin(10π/180))^2) = 9.80003 m/s^2.

Now, we can plug this value into the equation for the new period: T' = 2π√(L/9.80003).

To find the time difference in a year, we can simply subtract the two periods: T' - T = 2π√(L/9.80003) - 2π√(L/9.8).

This will give us the time difference in seconds. To convert it to years, we can divide by the number of seconds in a year (31,536,000).

I hope this helps! Let me know if you have any further questions.