A mass M,attached to a horizontal spring,exec | vedclass

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(A) At the mean position,the velocity of mass $M$ is maximum,given by $v_1 = A_1 \omega_1$,where $\omega_1 = \sqrt{\frac{K}{M}}$.When mass $m$ is plac

Vedclass · Accepted Answer

$\left( \frac{M + m}{M} \right)^{\frac{1}{2}}$

Vedclass · Answer

(A) At the mean position,the velocity of mass $M$ is maximum,given by $v_1 = A_1 \omega_1$,where $\omega_1 = \sqrt{\frac{K}{M}}$.When mass $m$ is placed on $M$,the momentum is conserved during the collision (as the impulsive force acts vertically,not horizontally).$M v_1 = (M + m) v_2$Substituting $v_1 = A_1 \sqrt{\frac{K}{M}}$ and $v_2 = A_2 \omega_2 = A_2 \sqrt{\frac{K}{M+m}}$:$M \left( A_1 \sqrt{\frac{K}{M}} \right) = (M + m) \left( A_2 \sqrt{\frac{K}{M + m}} \right)$$A_1 \sqrt{MK} = A_2 \sqrt{(M + m)K}$$\frac{A_1}{A_2} = \sqrt{\frac{M + m}{M}}$

$A$ mass $M$,attached to a horizontal spring,executes $S.H.M.$ with amplitude $A_1$. When the mass $M$ passes through its mean position,a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\left( \frac{A_1}{A_2} \right)$ is

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