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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) At the mean position,the potential energy is zero and the kinetic energy is maximum. Since no external horizontal force acts on the system during

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) At the mean position,the potential energy is zero and the kinetic energy is maximum. Since no external horizontal force acts on the system during the placement of mass $m$,linear momentum is conserved.Let $v_1$ be the velocity of mass $M$ at the mean position,and $v_2$ be the velocity of the combined mass $(M+m)$ immediately after placing $m$.Conservation of momentum: $M v_1 = (M + m) v_2$.The maximum velocity in $S.H.M.$ is given by $v_{max} = A \omega = A \sqrt{\frac{k}{m_{eff}}}$.For the first case: $v_1 = A_1 \sqrt{\frac{k}{M}}$.For the second case: $v_2 = A_2 \sqrt{\frac{k}{M+m}}$.Substituting these into the momentum equation:$M \left( A_1 \sqrt{\frac{k}{M}} \right) = (M + m) \left( A_2 \sqrt{\frac{k}{M+m}} \right)$.$A_1 \sqrt{M k} = A_2 \sqrt{(M+m) k}$.$A_1 \sqrt{M} = A_2 \sqrt{M+m}$.Therefore,$\frac{A_1}{A_2} = \sqrt{\frac{M+m}{M}} = \left( \frac{M+m}{M} \right)^{\frac{1}{2}}$.

$A$ 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 $\frac{A_1}{A_2}$ is

Similar Questions

All the springs in figures $(a)$, $(b)$, and $(c)$ are identical, each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$, and $T_c$ are the periodic times of oscillation of the three systems in figures $(a)$, $(b)$, and $(c)$ respectively, then:

$A$ $10 \,kg$ metal block is attached to a spring of spring constant $1000 \,N \,m^{-1}$. The block is displaced from the equilibrium position by $10 \,cm$ and released. The maximum acceleration of the block is: (in $\,m/s^2$)

The potential energy of a particle of mass $0.1\,kg,$ moving along the $x-$axis,is given by $U = 5x(x-4)\,J,$ where $x$ is in metres. It can be concluded that

On a smooth inclined plane,a block of mass $M$ is fixed to two rigid supports using two springs,as shown in the figure. If each spring has spring constant $k$,then the period of oscillation of the block is (Neglect the masses of the springs)

Vedclass Test Series

Exam Paper Generator

Online Exam Module