NTA Abhyas JEE Main2020PhysicsOscillationsPractice
A particle at the end of a spring executes simple harmonic motion with a period t 1 , while the corresponding period for another spring is t 2 . If the period of oscillation with the two springs in series is T , then -
Options
- AT = t 1 + t 2
- BT 2 = t 1 2 + t 2 2
- CT - 1 = t 1 - 1 + t 2 - 1
- DT - 2 = t 1 - 2 + t 2 - 2
Correct answer
B. T 2 = t 1 2 + t 2 2
Step-by-step solution
t 1 = 2 π m k 1   … . i ,   t 2 = 2 π m k 2   … . i i when springs are in series then T = 2 π m k 1 k 2 k 1 + k 2 = 2 π m k 1 + k 2 k 1 k 2 squaring and adding (i) and (ii) we get t 1 2 + t 2 2 = 4 π 2 m k 1 + 4 π 2 m k 2 = 4 π 2 m   k 1 + k 2 k 1 k 2 or t 1 2 + t 2 2 = T 2