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Question

(A) Let the displacement from the mean position be $x_1$ when velocity is $u$ and acceleration is $\alpha$. Since acceleration is $\alpha = \omega^2 x

Vedclass · Accepted Answer

$\frac{u^2-v^2}{\alpha+\beta}$

Vedclass · Answer

(A) Let the displacement from the mean position be $x_1$ when velocity is $u$ and acceleration is $\alpha$. Since acceleration is $\alpha = \omega^2 x_1$,we have $x_1 = \frac{\alpha}{\omega^2}$.At this position,$u^2 = \omega^2(A^2 - x_1^2) = \omega^2 A^2 - \omega^2 x_1^2 = \omega^2 A^2 - \alpha x_1$.Let the displacement be $x_2$ when velocity is $v$ and acceleration is $\beta$. Since acceleration is $\beta = \omega^2 x_2$,we have $x_2 = \frac{\beta}{\omega^2}$.At this position,$v^2 = \omega^2(A^2 - x_2^2) = \omega^2 A^2 - \omega^2 x_2^2 = \omega^2 A^2 - \beta x_2$.Subtracting the two velocity equations: $u^2 - v^2 = (\omega^2 A^2 - \alpha x_1) - (\omega^2 A^2 - \beta x_2) = \beta x_2 - \alpha x_1$.Substituting $x_1 = \frac{\alpha}{\omega^2}$ and $x_2 = \frac{\beta}{\omega^2}$:$u^2 - v^2 = \beta(\frac{\beta}{\omega^2}) - \alpha(\frac{\alpha}{\omega^2}) = \frac{\beta^2 - \alpha^2}{\omega^2}$.Thus,$\omega^2 = \frac{\beta^2 - \alpha^2}{u^2 - v^2}$.The distance between the two positions is $|x_2 - x_1| = |\frac{\beta}{\omega^2} - \frac{\alpha}{\omega^2}| = \frac{\beta - \alpha}{\omega^2}$.Substituting $\omega^2$: $|x_2 - x_1| = (\beta - \alpha) \cdot \frac{u^2 - v^2}{\beta^2 - \alpha^2} = \frac{(\beta - \alpha)(u^2 - v^2)}{(\beta - \alpha)(\beta + \alpha)} = \frac{u^2 - v^2}{\alpha + \beta}$.

List-$I$	List-$II$
$(1)$ Planck's constant	$(i)$ $[ML^{-1} T^{-2}]$
$(2)$ Gravitational constant	(ii) $[ML^{-1} T^{-1}]$
$(3)$ Bulk modulus	(iii) $[ML^2 T^{-1}]$
$(4)$ Coefficient of viscosity	(iv) $[M^{-1} L^3 T^{-2}]$

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