A mass m_1 connected to a horizontal spring p | vedclass

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

Question

(A) At the mean position,the potential energy is zero and the kinetic energy is maximum. The velocity $v$ of mass $m_1$ at the mean position is $v = A

Vedclass · Accepted Answer

$\left[\frac{m_1}{m_1 + m_2}\right]^{\frac{1}{2}}$

Vedclass · Answer

(A) At the mean position,the potential energy is zero and the kinetic energy is maximum. The velocity $v$ of mass $m_1$ at the mean position is $v = A\omega = A\sqrt{\frac{k}{m_1}}$.When mass $m_2$ is placed on $m_1$,the momentum is conserved because the spring force is zero at the mean position. Let $v'$ be the new velocity:$m_1 v = (m_1 + m_2) v'$$v' = \frac{m_1 v}{m_1 + m_2} = \frac{m_1 A \sqrt{k/m_1}}{m_1 + m_2} = A \sqrt{\frac{k m_1}{(m_1 + m_2)^2}}$.The new energy of the system is $E' = \frac{1}{2} k A_1^2 = \frac{1}{2} (m_1 + m_2) (v')^2$.Substituting $v'$:$\frac{1}{2} k A_1^2 = \frac{1}{2} (m_1 + m_2) \left( A^2 \frac{k m_1}{(m_1 + m_2)^2} \right)$.$A_1^2 = A^2 \frac{m_1}{m_1 + m_2}$.Therefore,$\frac{A_1}{A} = \sqrt{\frac{m_1}{m_1 + m_2}}$.

$A$ mass $m_1$ connected to a horizontal spring performs $S.H.M.$ with amplitude $A$. While mass $m_1$ is passing through the mean position,another mass $m_2$ is placed on it so that both the masses move together with amplitude $A_1$. The ratio of $\frac{A_1}{A}$ is:

Similar Questions

Find the maximum amplitude for safe $SHM$ (the block does not topple during $SHM$) of a cubical block of side $a$ on a smooth horizontal floor as shown in the figure (the spring is massless). The spring is attached at a height of $2a/3$ from the floor.

$A$ body of mass $m$ is suspended to an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

$A$ spring has a certain mass suspended from it and its period for vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio $T_1 / T_2$ is

Find the ratio of time periods of two identical springs if they are first joined in series and then in parallel and a mass $m$ is suspended from them.

$A$ particle is suspended from a vertical spring which is executing $S.H.M.$ of frequency $5 \ Hz$. The spring is unstretched at the highest point of oscillation. What is the maximum speed of the particle? (Take $g = 10 \ m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module