Define simple pendulum and derive it's expestion time period
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You need to remember the formula of constant of simple pendulum such that:
k = L/T^2
You need to consider the value of constant equivalent to g/(4pi^2) (g expresses the gravity acceleration)
You need to set the equations g/(4pi^2) and L/T^2 equal such that:
an ideal pendulum consisting of a point mass suspended by a weightless inextensible perfectly flexible thread and free to vibrate without friction
derivation==--
L/T^2 = g/(4pi^2)
You need to find time period such that:
g*T^2 = 4pi^2*L
T^2 = (4pi^2*L)/g => T = 2pi*sqrt(L/g)
Hence, evaluating the time period of simple pendulum under given conditions yields T = 2pi*sqrt(L/g).
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To and Fro motion of the pendulum in two second is called Second pendulum
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