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(B) In $SHM$,the velocity $V$ of a particle at a distance $x$ from the mean position is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplit

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{{{x_2}^2 - {x_1}^2}}{{{V_1}^2 - {V_2}^2}}}$

Vedclass · Answer

(B) In $SHM$,the velocity $V$ of a particle at a distance $x$ from the mean position is given by $V = \omega \sqrt{a^2 - x^2}$,where $a$ is the amplitude and $\omega$ is the angular frequency.Squaring both sides,we get $V^2 = \omega^2(a^2 - x^2)$.For distances $x_1$ and $x_2$,we have:$V_1^2 = \omega^2(a^2 - x_1^2) \dots (i)$$V_2^2 = \omega^2(a^2 - x_2^2) \dots (ii)$Subtracting equation $(ii)$ from $(i)$:$V_1^2 - V_2^2 = \omega^2(a^2 - x_1^2 - a^2 + x_2^2)$$V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2)$$\omega^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$Since the time period $T = \frac{2\pi}{\omega}$,we have $\omega = \frac{2\pi}{T}$.Substituting this into the equation for $\omega^2$:$\left(\frac{2\pi}{T}\right)^2 = \frac{V_1^2 - V_2^2}{x_2^2 - x_1^2}$$T^2 = 4\pi^2 \left(\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}\right)$$T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}$

$A$ 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

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