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=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\\alpha=0.3\/3=0.1\\to \\alpha\\approx5.7\u00b0"

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 "l_1=l+0.2l=1.2l"

We get that

"\\frac{T}{T_1}=\\frac{1}{\\sqrt{1.2}}\\to T_1=\\sqrt{1.2}T=\\sqrt{1.2}\\cdot3.50=3.8s"

Learn more about our help with Assignments: MechanicsRelativity

## Comments

## Leave a comment