How long must an ideal pendulum (or the radius of an air track) be such that one swing (half period) lasts very near one second? Use = 9.81 m/s
How long must an ideal pendulum be such that one swing (half period) lasts very near one second when it is placed on Mars? The gravity on Mars' surface is g = 3.72 m/s .
To find the length of an ideal pendulum for a half period of one second, we can use the formula for the period of a pendulum:
T = 2π√(L/g)
Where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
For the first question, on Earth where g = 9.81 m/s², we can rearrange the formula to solve for L:
L = (T/(2π))² * g
Since we want the half period to equal one second (T = 1 second), we can substitute this value into the equation and calculate the length:
L = (1 / (2π))² * 9.81
L = 0.0794 meters (or 79.4 cm)
Therefore, the ideal length of the pendulum on Earth for a half period of one second is approximately 79.4 cm.
For the second question, on Mars where g = 3.72 m/s², we can use the same formula and substitute the value of g for Mars:
L = (1 / (2π))² * 3.72
L = 0.0308 meters (or 30.8 cm)
Therefore, the ideal length of the pendulum on Mars for a half period of one second is approximately 30.8 cm.