The period of a pendulum is give approximately by the formula
t = 2*pi*sqrt(l/g) where l is the length of the pendulum and g is the acceleration (not accerlation) due to gravity. Thus g is part of the formula for the period.
Yes. Given a constant for gravity, the period of the pendulum is a function of it's length to the center of mass. In a higher gravity, the period would be shorter for the same length of pendulum.
Height does not affect the period of a pendulum.
The period of a pendulum (in seconds) is 2(pi)√(L/g), where L is the length and g is the acceleration due to gravity. As acceleration due to gravity increases, the period decreases, so the smaller the acceleration due to gravity, the longer the period of the pendulum.
It messes up the math. For large amplitude swings, the simple relation that the period of a pendulum is directly proportional to the square root of the length of the pendulum (only, assuming constant gravity) no longer holds. Specifically, the period increases with increasing amplitude.
For a simple pendulum, consisting of a heavy mass suspended by a string with virtually no mass, and a small angle of oscillation, only the length of the pendulum and the force of gravity affect its period. t = 2*pi*sqrt(l/g) where t = time, l = length and g = acceleration due to gravity.
Yes. Given a constant for gravity, the period of the pendulum is a function of it's length to the center of mass. In a higher gravity, the period would be shorter for the same length of pendulum.
Height does not affect the period of a pendulum.
In an ideal pendulum, the only factors that affect the period of a pendulum are its length and the acceleration due to gravity. The latter, although often taken to be constant, can vary by as much as 5% between sites. In a real pendulum, the amplitude will also have an effect; but if the amplitude is relatively small, this can safely be ignored.
The period of a pendulum (in seconds) is 2(pi)√(L/g), where L is the length and g is the acceleration due to gravity. As acceleration due to gravity increases, the period decreases, so the smaller the acceleration due to gravity, the longer the period of the pendulum.
It messes up the math. For large amplitude swings, the simple relation that the period of a pendulum is directly proportional to the square root of the length of the pendulum (only, assuming constant gravity) no longer holds. Specifically, the period increases with increasing amplitude.
For a simple pendulum, consisting of a heavy mass suspended by a string with virtually no mass, and a small angle of oscillation, only the length of the pendulum and the force of gravity affect its period. t = 2*pi*sqrt(l/g) where t = time, l = length and g = acceleration due to gravity.
Normally the acceleration of gravity is not a factor in the period of a simple pendulum because it does not change on Earth, but if it were to be put on another celestial body the period would change. As gravity increases the period is shorter and as the gravity is less the period is longer.
In a simple pendulum, with its entire mass concentrated at the end of a string, the period depends on the distance of the mass from the pivot point. A physical pendulum's period is affected by the distance of the centre-of-gravity of the pendulum arm to the pivot point, its mass and its moment of inertia about the pivot point. In real life the pendulum period can also be affected by air resistance, temperature changes etc.
The period increases as the square root of the length.
They determine the length of time of the pendulum's swing ... its 'period'.
It doesn't. Period depends on the length of the pendulum and the acceleration of gravity. Adding weight doesn't change the period at all.
The period of a simple pendulum swinging at a small angle is approximately 2*pi*Sqrt(L/g), where L is the length of the pendulum, and g is acceleration due to gravity. Since gravity on the moon is approximately 1/6 of Earth's gravity, the period of a pendulum on the moon with the same length will be approximately 2.45 times of the same pendulum on the Earth (that's square root of 6).