A particle executes SHM. Its velocities are v | vedclass

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(B) The velocity $v$ of a particle in $SHM$ at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude and $\omega$ is t

Vedclass · Accepted Answer

$\frac{1}{2\pi} \left[ \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \right]^{1/2}$

Vedclass · Answer

(B) The velocity $v$ of a particle in $SHM$ at displacement $x$ 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 $v^2 = \omega^2(A^2 - x^2) = \omega^2 A^2 - \omega^2 x^2$.For the given conditions:$v_1^2 = \omega^2 A^2 - \omega^2 x_1^2$  --- $(1)$$v_2^2 = \omega^2 A^2 - \omega^2 x_2^2$  --- $(2)$Subtracting equation $(2)$ from $(1)$:$v_1^2 - v_2^2 = \omega^2(x_2^2 - x_1^2)$$\omega^2 = \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2}$$\omega = \left[ \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \right]^{1/2}$Since frequency $f = \frac{\omega}{2\pi}$,we have:$f = \frac{1}{2\pi} \left[ \frac{v_1^2 - v_2^2}{x_2^2 - x_1^2} \right]^{1/2}$

$A$ particle executes $SHM$. Its velocities are $v_1$ and $v_2$ at displacements $x_1$ and $x_2$ from the mean position,respectively. The frequency of oscillation will be

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