how to find the answer

Question

Two simple pendulums of length 1 m and 16 m are in phase at the mean position at a certain instant of time. If T is the time period of shorter pendulum then the minimum time after which they will again in phase is

Answer 1

 
how to find the answer
Two simple pendulums of length 1 m and 16 m are in phase at the mean position at a certain instant of time. If T is the time period of shorter pendulum then the minimum time after which they will again in phase is

 

Solution:

 

the formula definition for the period of simple pendulum

is T = 2π√l/g = 2π(√l/√g)=(2π/√g)(√l)

 

in the equation above (2π/√g) is the same for both pendulums

 

The period of the 1 m long pendulum T₁ = (2π/√g)√1=(2π/√g)

The period of the 16 m long pendulum T₁₆ = (2π/√g)√16=4(2π/√g)

 

They start in phase at the same position at T

then T1/T₁₆=1/4, and T₁₆= 4T₁ This means that the 1 m pendulum completes four periods before the 16 m pendulum completes one period

 

In other words the 1 m pendulum completes 4 full oscillations when the 16 m pendulum completes only one. Therefore they will be in phase again at the same point after time T = 4(2π/√g)=4(2π)/√9.81m/s²

 

T=8π/3.13= 8.0 sec