A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2 respectively. Its time period is
- 2 V12+V22x12+x22
- 2 V12-V22x12-x22
- 2 x12+x22V12+V22
- 2 x22-x12V12-V22
Answer
(4) 2 x22-x12V12-V22
Solution
In SHM: V2 = 2(A2 - x2)
V12 = 2(A2 - x12) and V22 = 2(A2 - x22)
Subtracting: V12 - V22 = 2(x22 - x12)
= V12 - V22x22 - x12
T = 2 = 2 x22 - x12V12 - V22
Answer: (4)