A light spring is suspended with mass m_1 at | vedclass

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

Question

(A) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant.For the fi

Vedclass · Accepted Answer

$4 : 5$

Vedclass · Answer

(A) The time period of a mass-spring system is given by $T = 2\pi \sqrt{\frac{m}{k}}$, where $m$ is the mass and $k$ is the spring constant.For the first case, $T = 2\pi \sqrt{\frac{m_1}{k}}$.For the second case, the total mass is $(m_1 + m_2)$, so the new time period is $T' = 2\pi \sqrt{\frac{m_1 + m_2}{k}} = \frac{3}{2} T$.Dividing the two equations: $\frac{T'}{T} = \sqrt{\frac{m_1 + m_2}{m_1}} = \frac{3}{2}$.Squaring both sides: $\frac{m_1 + m_2}{m_1} = \frac{9}{4}$.This simplifies to $1 + \frac{m_2}{m_1} = \frac{9}{4}$, which means $\frac{m_2}{m_1} = \frac{9}{4} - 1 = \frac{5}{4}$.Therefore, the ratio $\frac{m_1}{m_2} = \frac{4}{5}$.

Similar Questions

Five identical springs are used in the following three configurations. The time periods of vertical oscillations in configurations $(i)$,$(ii)$ and $(iii)$ are in the ratio

$A$ $5\, kg$ collar is attached to a spring of spring constant $500\, N/m$. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by $10\, cm$ and released. The time period of oscillation is

$A$ spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

Define the torsional constant for a spring.

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