A science museum has asked you to design a simple pendulum that will make 25.0 complete swings in 85.0 {\rm s}.
How long should the pendulum be? in meters
T =85/25 = 3.4 seconds per swing
T/2pi = .541 = sqrt(L/9.8)
L/9.8 = .293
L = 2.87 m
A science museum has asked you to design a simple pendulum that will make 54 complete swings in 73 s. What length should you specify for this pendulum?
To determine the length of the pendulum, we can use the formula for the period of a simple pendulum:
T = 2π * sqrt(L/g)
Where:
T is the period (time per swing) in seconds,
L is the length of the pendulum in meters,
and g is the acceleration due to gravity, approximately 9.8 m/s^2 on Earth.
Given that the pendulum completes 25 complete swings in 85.0 s, we can calculate its period:
T = 85.0 s / 25 = 3.4 s
Now, rearranging the formula above to solve for L:
L = (T^2 * g) / (4π^2)
Substituting the values:
L = (3.4 s)^2 * 9.8 m/s^2 / (4π^2)
L = 1.156 m (rounded to three significant figures)
Therefore, the length of the pendulum should be approximately 1.16 meters.
No book handy so will derive pendulum
let theta be angle from vertical then
(1/2) m v^2 max = m g h = m g L (1- (1-theta^2/2)) maximum
(1/2) v^2 = g L(theta^2/2)
but v = L d theta/dt
L^2 d (theta/dt)^2 = g L theta^2
let Theta = A sin 2 pi t/T
max = A
then d theta/dt = A (2 pi/T) cos 2pit/T
max = A(2 pi /T)
so
L (2 pi/T)^2 = g
2 pi/T = sqrt(g/L)
T = 2 pi sqrt(L/g)
so now the problem
T =85/25 = 3.4 seconds per swing
2 pi/T = 1.85
T^2 = 3.42 = L/9.81
L = 33.5 m