Question

4. Compare the time period of the two simple pendulums of length 1 m and 16 m at a place.​

Answer 1

Answer:T1=2π1g−−√,T2=2πu16g−−−√=4T1  

at any time t phase of pendulums are

ϕ1=ω1t=2πT1t,ϕ2=ω2t=2πT2t

First pendulum is faster both will be in same phase again when faster pendulum completated one oscillation more than slower pendulum

⇒2πT1t−2π4T1t=2π⇒t=4T13

No of oscillation completed by shorter pendulum in time t

n=tT1=43

Explanation: