SHMHard
Question
Two identical springs are fixed at one end and masses 1 kg and 4kg are suspended at their other ends. They are both stretched down from their mean position and let go simultaneously. If are in the same phase after energy 4 seconds then the spring constant k is
Options
A.
B.
C.
D.given data is insufficient
Solution
ω1 = 
, ω2 = 
(ω1 - ω2)t = 0, 2π, 4π............t = 0
⇒ t =
= 4 ⇒ 2π = 4 (ω1 - ω2)
⇒
= (ω1 - ω2) ⇒ k = π2 N/m
, ω2 = (ω1 - ω2)t = 0, 2π, 4π............t = 0
⇒ t =
= 4 ⇒ 2π = 4 (ω1 - ω2)⇒
= (ω1 - ω2) ⇒ k = π2 N/mCreate a free account to view solution
View Solution FreeMore SHM Questions
On loading a spring with bob, its period of oscillation in a vertical plane is T. If this spring pendulum is tied with o...The mass of particle executing S.H.M. is 1 gm. If its periodic time is π seconds, the value of force constant is :...Two particle executes S.H.M of same amplitude and frequency along the same straight line. They pass one another when goi...The displacement - time graph of particle executing SHM is shown. Which of the following statements is / are true?...If amplitude of the particle which is executing S.H.M., is doubled, then which quantity will become double ?...