Q117 / 180 · NEET 2015 (Phase 1) · Physics

A particle is executing SHM along a straight line. Its velocities at distances $x_{1}$ and $x_{2}$ from the mean position are $V_{1}$ and $V_{2}$ respectively. Its time period is:

A
$2 π \sqrt{\frac{x_{1}^{2} + x_{2}^{2}}{V_{1}^{2} + V_{2}^{2}}}$
Correct answer
B
$2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}$
C
$2 π \sqrt{\frac{V_{1}^{2} + V_{2}^{2}}{x_{1}^{2} + x_{2}^{2}}}$
D
$2 π \sqrt{\frac{V_{1}^{2} - V_{2}^{2}}{x_{1}^{2} - x_{2}^{2}}}$

Solution

For particle undergoing SHM,

V = ω \sqrt{A^{2} - x^{2}} \Rightarrow V^{2} = ω^{2} (A^{2} - x^{2})

V_{1} = ω \sqrt{A^{2} - x_{1}^{2}} \Rightarrow V_{1}^{2} = ω^{2} (A^{2} - x_{1}^{2})

...(i)

V_{2} = ω \sqrt{A^{2} - x_{2}^{2}} \Rightarrow V_{2}^{2} = ω^{2} (A^{2} - x_{2}^{2})

...(ii)

V_{1}^{2} - V_{2}^{2} = ω^{2} (x_{2}^{2} - x_{1}^{2})

ω = \sqrt{\frac{V_{1}^{2} - V_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}}

T = 2 π \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{V_{1}^{2} - V_{2}^{2}}}

Hence, the correct answer is option: 'B'.