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The pendulum frequency is dependent upon the length of the pendulum. The torque is the turning force of the pendulum.
multiply *2 of the previous period
It doesn't. Only the length of the pendulum and the strength of the gravitational field alter the period/frequency.
-- its length (from the pivot to the center of mass of the swinging part) -- the local acceleration of gravity in the place where the pendulum is swinging
It's not always the same. The frequency of a pendulum depends on its length, on gravity, on the pendulum's exact shape, and on the amplitude. For a small amplitude, and for a pendulum that has all of its mass concentrated in one point, the period is 2 x pi x square root of (L / g) (where L=length, g=gravity). The frequency, of course, is the reciprocal of this.
time period of simple pendulum is dirctly proportional to sqare root of length...
The frequency of a pendulum varies with the square of the length.
The frequency of a pendulum is inversely proportional to the square root of its length.
A longer pendulum will have a smaller frequency than a shorter pendulum.
The period of the pendulum is (somewhat) inversely proportional to the square root of the length. Therefore, the frequency, the inverse of the period, is (somewhat) proportional to the square root of the length.
For relatively small oscillations, the frequency of a pendulum is inversely proportional to the square root of its length.
The pendulum frequency is dependent upon the length of the pendulum. The torque is the turning force of the pendulum.
The period increases - by a factor of sqrt(2).
multiply *2 of the previous period
It doesn't. Only the length of the pendulum and the strength of the gravitational field alter the period/frequency.
Let us go step by step Period = 2 pi ./l/g Or frequency = 1/2pi * ./g/l Or 2 pi frequency = angular frequency = ./g/l As we reduce the length by 4 times i.e 1/4 l then we have angular frequency doubled. Hence reduce the length to 0.25 l
9.7m