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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
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2π 
x22 - x12 
1 / 2 V12 V22 -
2π V12 + V22 1 / 2 x12 + x22 -
2π V12 V22 1 / 2 x12 - x22 -
2π x12 - x22 1 / 2 V12 - V22
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Correct Option: A
As we know, for particle undergoing SHM,
V = ω √A² - X²
V12 = ω2 (A2 - x12)
V22 = ω2 (A2 - x22)
Substracting we get,
| + x12 = | + x22 | |||
| ω2 | ω2 |
| ⇒ | = x22 - x12 | |
| ω2 |
| ⇒ w = |  | |
| x22 - x12 |
| ⇒ T = 2π | | |
| V12 - V22 |
