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For a simple pendulum, with very small, technically infinitesimal, swings, the period T is approximately 2 pisquare root (Length / gravity). From there, it is simple algebra.
T2 = 4 pi2 length / gravity
gravity = 4 pi2 length / T2
For more information, including how to compensate for circular error introduced by non-infinitesimal swings, please see the Related Link below.
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What is the maths of period of pendulum?
For small swings, and a simple pendulum:T = 2 pi root(L/g) where T is the time for one period, L is the length of the pendulum, and g is the strength of the gravitational field.
What is the relationship between the period of a pendulum and its length?
For small angles, the formula for a pendulum's period (T) can be approximated by the formula:T = 2 * pi * sqrt(L/g), where L is the length of the pendulum length, and g is acceleration due to gravity. See related link for Simple Pendulum.
How do you solve a problem in numerical analysis?
g
What is the difference in period for a pendulum on earth and a pendulum on moon?
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).
What is the effect of acceleration due to gravity on the time period of a simple pendulum?
T=2pi(l/g)1/2
Value of g by simple pendulum?
The acceleration due to gravity (g) can be determined using a simple pendulum by measuring the period of oscillation and using the formula T = 2Οβ(L/g), where T is the period of oscillation, L is the length of the pendulum, and g is the acceleration due to gravity. By rearranging the formula, we can solve for g as g = (4Ο^2*L) / T^2.
How long must a simple pendulum be in order to have a period of one second?
A simple pendulum must be approximately 0.25 meters long to have a period of one second. This length is calculated using the formula for the period of a simple pendulum, which is T = 2Οβ(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. By substituting T = 1 second and g = 9.81 m/s^2, you can solve for L.
What is the equation for the period of a simple pendulum?
The equation for the period (T) of a simple pendulum is T = 2Οβ(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.
What factors determine the time period of the simple pendulum?
The time period of a simple pendulum is determined by the length of the pendulum, the acceleration due to gravity, and the angle at which the pendulum is released. The formula for the time period of a simple pendulum is T = 2Οβ(L/g), where T is the time period, L is the length of the pendulum, and g is the acceleration due to gravity.
Would be value of g be affected if the size of the Bob is varied in simple pendulum experiment?
No, the value of acceleration due to gravity (g) would not be affected by changing the size of the bob in a simple pendulum experiment. The period of a simple pendulum is determined by the length of the pendulum and the gravitational acceleration at that location, not the size of the bob.
What happen to the frequency of a simple pendulum when its length is doubled?
When the length of a simple pendulum is doubled, the frequency of the pendulum decreases by a factor of β2. This relationship is described by the formula T = 2Οβ(L/g), where T is the period of the pendulum, L is the length, and g is the acceleration due to gravity.
What is the maths of period of pendulum?
For small swings, and a simple pendulum:T = 2 pi root(L/g) where T is the time for one period, L is the length of the pendulum, and g is the strength of the gravitational field.
How would the time period of a simple pendulum clock be affected if it were on the moon instead of the earth?
The time period of a pendulum would increases it the pendulum were on the moon instead of the earth. The period of a simple pendulum is equal to 2*pi*√(L/g), where g is acceleration due to gravity. As gravity decreases, g decreases. Since the value of g would be smaller on the moon, the period of the pendulum would increase. The value of g on Earth is 9.8 m/s2, whereas the value of g on the moon is 1.624 m/s2. This makes the period of a pendulum on the moon about 2.47 times longer than the period would be on Earth.
What is the relationship between the period of a pendulum and its length?
For small angles, the formula for a pendulum's period (T) can be approximated by the formula:T = 2 * pi * sqrt(L/g), where L is the length of the pendulum length, and g is acceleration due to gravity. See related link for Simple Pendulum.
How do you calculate the acceleration due to gravity using a simple pendulum?
The acceleration due to gravity can be calculated using a simple pendulum by measuring the period of oscillation (time taken for the pendulum to complete one full swing) and the length of the pendulum. The formula to calculate acceleration due to gravity is: g = 4ΟΒ²L / TΒ², where g is acceleration due to gravity, L is the length of the pendulum, and T is the period of oscillation.
How do you solve a problem in numerical analysis?
g
How do you solve pendulum to frequency in vibrations?
To determine the frequency of a pendulum's vibrations, you can use the formula: frequency = 1 / (2 * pi) * sqrt(g / L), where g is the acceleration due to gravity (9.81 m/s^2) and L is the length of the pendulum. Plug in the values for g and L into the formula to calculate the frequency in hertz.
