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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?
You can build a simple pendulum - one that has most of its mass concentrated in a small place, at the end of the pendulum. Measure the pendulum's length, and measure how long it takes to go back and forth. Use the formula for the period of a pendulum, solving for "g".
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.
Does amplitude effect the period of a pendulum?
no it doesnt affect the period of pendulum. the formulea that we know for simple pendulum is T = 2pie root (L/g)
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 solve a problem in numerical analysis?
g
Does the mass of the pendulum bob affect the value of g why?
The value of gravitational acceleration 'g' is totally unaffected by changing mass of the body. We are not talking about weight of the pendulum. It is the value 'g' we are talking about, which remains unaffected by changing mass as: g= ((2xpie)2)xL)/T2 where, g= gravitational acceleration L= length of simple pendulum T= time period in which the pendulum completes its single vibration or oscillation
How do you calculate the acceleration due to gravity using a simple pendulum?
The period of a simple pendulum is 2 pi (L/g)1/2. Construct a pendulum and set it into motion. Measure the period for small swings. Back-calculate g...t = 2 pi (L/g)1/2t2 = 4 pi2 L/gg = 4 pi2 L/t2
What is the period of a pedelum?
The period of oscillation of a simple pendulum displaced by a small angle is: T = (2*PI) * SquareRoot(L/g) where T is the period in seconds, L is the length of the string, and g is the gravitional field strength = 9.81 N/Kg. This equation is for a simple pendulum only. A simple pendulum is an idealised pendulum consisting of a point mass at the end of an inextensible, massless, frictionless string. You can use the simple pendulum model for any pendulum whose bob mass is much geater than the length of the string. For a physical (or real) pendulum: T = (2*PI) * SquareRoot( I/(mgr) ) where I is the moment of inertia, m is the mass of the centre of mass, g is the gravitational field strength and r is distance to the pivot from the centre of mass. This equation is for a pendulum whose mass is distributed not just at the bob, but throughout the pendulum. For example, a swinging plank of wood. If the pendulum resembles a point mass on the end of a string, then use the first equation.
What is the problem that G shock can solve?
G-Shock is the name given to a brand of watches that are manufactured and made by the Cascio company. The problem that these watches can solve is the question of what time it is.
What the time periods of a simple pendulum at equator of the earth?
For small swings of a mass suspended on a weightless string, the period is given by T = 2 pi sqrt (a/g) where a is the length of the pendulum and g is the acceleration due to gravity.
What is the time period of simple pendulum at center of earth?
i think it is infinite because acceleration due to gravity at the center of the earth is zero and time period of the simple pendulum is given by 2*3.14*sqrt(l/g)....
