Question #125837

A simple pendulum is swinging with an amplitude of 30cm and a period of 3.50s. what will be the pendulum's period if it's length is extended by 20% the mass is doubled, and the swing has an amplitude of 20cm.

Expert's answer

T=2πlg→l=T2⋅g4π2=3.52⋅9.814⋅3.142=3mT=2\pi\sqrt{\frac{l}{g}}\to l=\frac{T^2\cdot g}{4\pi^2}=\frac{3.5^2\cdot 9.81}{4\cdot3.14^2}=3m


So,


sin⁡α=0.3/3=0.1→α≈5.7°\sin\alpha=0.3/3=0.1\to \alpha\approx5.7°


The period of oscillation of a simple pendulum does not depend on the mass of the pendulum and the amplitude (if the angle of deviation of the pendulum does not exceed 6 °)


if the length is extended by 20% then l1=l+0.2l=1.2ll_1=l+0.2l=1.2l


We get that


TT1=11.2→T1=1.2T=1.2⋅3.50=3.8s\frac{T}{T_1}=\frac{1}{\sqrt{1.2}}\to T_1=\sqrt{1.2}T=\sqrt{1.2}\cdot3.50=3.8s










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