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Question

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
2πV21+V22x21+x22
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B
2πV21V22x21x22
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C
2πx21+x22V21+V22
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D
2πx22x21V21V22
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Solution

The correct option is D $$2\pi \sqrt {\cfrac {x_2^2-x_1^2}{V_1^2-V_2^2}}$$

Using the expressions for velocity for SHM,

$$v_1 = \omega \sqrt{A^2 -x_1^2}$$
$$v_2 = \omega \sqrt{A^2 -x_2^2}$$
Squaring both,
$$v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)$$ ....(3)
$$v_2 ^2 = \omega ^2 ( A^2 -x_2 ^2)$$ ....(4)
Subtracting (4) from (3), $$v_1 ^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2) $$
$$ \Rightarrow  4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )$$
$$ \Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2 ^2}}$$ where we have used $$\omega = \dfrac{2\pi}{T}$$


Physics

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