A grandfather clock is constructed so that it has a simple pendulum that swings from one side to the other, a distance of 23.0 mm, in 1.00 s.
A)What is the maximum speed of the pendulum bob? Use two different methods. First, assume SHM and use the relationship between amplitude and maximum speed.(in cm/s)
B) Second, use energy conservation. Assume g = 9.81 m/s2
*** The problem is I am not getting the right answer and I find it oddly that A and B is 3.61 cm/s. It would be great if so one could look over it for me. Check for my equations and my calculation,I'm notorious for screwing up calculations and miss using equations. Please help thanks.
A)For a pendulum omega (angular frequency) = 2*pi/T where T is the period Here T = 2.0s (time for one complete oscillation) so omega = pi = 3.14 rad/s. From the eqns for SHM you can derive the function for vmax.... vmax = omega*A where A is the amplitude (In this problem A = 0.023/2m)
B) Here (K +U)bottom = (K + U) top Note U bot = 0 and K top = 0
So K bot =1/2*m*vmax^2 = m*g*y or vmax = sqrt(2*g*y) All we need is y. The length of a pendulum with a period of 2.0s is T^2*g/(4*pi^2) = 0.993m
So the angle between the bottom and the top can be found using s = R*theta theta = s/R
The Attempt at a Solution
A)So vmax = 3.14*0.023/2 = 0.0361 m/s = 3.61cm/s
B)Therefore the height of the bob is 0.993*(1-cos(.6635)) = 6.658x10^-5 m
So vmax = sqrt(2*6.658x10^-5*9.8) = 0.0361m/s = 3.61cm/s
Any help is great just need someone to look over my work because im not getting the right answer.