Question
5. The length of a simple pendulum with a period on Earth of one second is most nearly
a.) 0.12 m
b.) 0.25 m
c.) 0.50 m
d.) 1.0 m
e.) 10.0 m
The formula for the period is
P = 2 pi sqrt(L/g)
g is the acceleration of gravity, 9.8 m/s^2.
Solve for L. (or get out a piece of string and do the experiment.)
a.) 0.12 m
b.) 0.25 m
c.) 0.50 m
d.) 1.0 m
e.) 10.0 m
The formula for the period is
P = 2 pi sqrt(L/g)
g is the acceleration of gravity, 9.8 m/s^2.
Solve for L. (or get out a piece of string and do the experiment.)
Answers
kirafreedom
Pendulum period in seconds
T ≈ 2π√(L/g)
L is length of pendulum in meters
g is gravitational acceleration = 9.8 m/s²
Length for 1 second = 0.248 m
(b)
T ≈ 2π√(L/g)
L is length of pendulum in meters
g is gravitational acceleration = 9.8 m/s²
Length for 1 second = 0.248 m
(b)