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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.
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
What is the equation for the period of a simple pendulum?
The equation for the period (T) of a simple pendulum is T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.
What is the formula for the angular frequency of a simple pendulum in terms of the acceleration due to gravity and the length of the pendulum?
The formula for the angular frequency () of a simple pendulum is (g / L), where g is the acceleration due to gravity and L is the length of the pendulum.
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 = 2π√(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.
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 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 equation for the period of a simple pendulum?
The equation for the period (T) of a simple pendulum is T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity.
What is the formula for the angular frequency of a simple pendulum in terms of the acceleration due to gravity and the length of the pendulum?
The formula for the angular frequency () of a simple pendulum is (g / L), where g is the acceleration due to gravity and L is the length of the pendulum.
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 = 2π√(L/g), where T is the time period, L is the length of the pendulum, and g is the acceleration due to gravity.
What is the formula for calculating the angular frequency of a simple pendulum?
The formula for calculating the angular frequency of a simple pendulum is (g / L), where represents the angular frequency, g is the acceleration due to gravity, and L is the length of the pendulum.
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 √2. This relationship is described by the formula T = 2π√(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 effect of changing length or mass of the pendulum on the value of g?
Changing the length or mass of a pendulum does not affect the value of acceleration due to gravity (g). The period of a pendulum depends on the length of the pendulum and not on its mass. The formula for the period of a pendulum is T = 2π√(L/g), where T is the period, L is the length of the pendulum, 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 you determine the value of gravity using a pendulum?
Set the pendulum swinging, with only a very small initial angular displacement. Measure the time taken to complete a certain number of oscillations, and then establish the average duration T of an oscillation. If the length of the pendulum is L, then gravitational field strength g is approximated by g = ~4pi2L/T2 This result derives from the modelling of the pendulum as a simple harmonic oscillator; for this to be a realistic model, the amplitude of oscillations must be small.
What is the period of a simple pendulum 45 cm long on the Earth?
For small amplitudes, the period can be calculated as 2 x pi x square root of (L / g). Convert the length to meters, and use 9.8 for gravity. The answer will be in seconds. About 1.4 seconds.
