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

2\pi \sqrt V_1^2 + V_2^2 \over x_1^2 + x_2^2

2\pi \sqrt V_1^2 - V_2^2 \over x_1^2 - x_2^2

2\pi \sqrt x_1^2 + x_2^2 \over V_1^2 + V_2^2

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Oscillations question

2015 · Q124

Oscillations question

2015 · Q124

View written solutionFree

Correct answer: D

As we know, for particle undergoing SHM,

$V = \omega \sqrt {{A^2} - {X^2}}$

$V_1^2 = {\omega ^2}\left( {{A^2} - x_1^2} \right)$

$V_2^2 = {\omega ^2}\left( {{A^2} - x_2^2} \right)$

Substracting we get,

${{V_1^2} \over {{\omega ^2}}} + x_1^2 = {{V_2^2} \over {{\omega ^2}}} + x_2^2$

$\Rightarrow {{V_1^2 - V_2^2} \over {{\omega ^2}}} = x_2^2 - x_1^2$

$\Rightarrow w = \sqrt {{{V_1^2 - V_2^2} \over {x_2^2 - x_1^2}}}$

$\Rightarrow T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {V_1^2 - V_2^2}}}$

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Oscillations question

2015 · Q124

Oscillations question

2015 · Q124

View written solutionFree

Correct answer: D

As we know, for particle undergoing SHM,

$V = \omega \sqrt {{A^2} - {X^2}}$

$V_1^2 = {\omega ^2}\left( {{A^2} - x_1^2} \right)$

$V_2^2 = {\omega ^2}\left( {{A^2} - x_2^2} \right)$

Substracting we get,

${{V_1^2} \over {{\omega ^2}}} + x_1^2 = {{V_2^2} \over {{\omega ^2}}} + x_2^2$

$\Rightarrow {{V_1^2 - V_2^2} \over {{\omega ^2}}} = x_2^2 - x_1^2$

$\Rightarrow w = \sqrt {{{V_1^2 - V_2^2} \over {x_2^2 - x_1^2}}}$

$\Rightarrow T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {V_1^2 - V_2^2}}}$