Two masses m_1 and m_2 connected by a spring | vedclass

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(A) The correct option is $A$.Let the displacements of masses $m_1$ and $m_2$ from their equilibrium positions be $x_1$ and $x_2$,respectively.The tot

Vedclass · Accepted Answer

$T=2 \pi \sqrt{\frac{1}{k}\left(\frac{m_1 m_2}{m_1+m_2}\right)}$

Vedclass · Answer

(A) The correct option is $A$.Let the displacements of masses $m_1$ and $m_2$ from their equilibrium positions be $x_1$ and $x_2$,respectively.The total elongation of the spring is $x = x_1 + x_2$.The restoring force on each mass is $F = -kx$.Applying Newton's second law to each mass:$m_1 \frac{d^2 x_1}{dt^2} = -kx$$m_2 \frac{d^2 x_2}{dt^2} = -kx$From these,we have:$\frac{d^2 x_1}{dt^2} = -\frac{k}{m_1} x$$\frac{d^2 x_2}{dt^2} = -\frac{k}{m_2} x$Since $x = x_1 + x_2$,the relative acceleration is:$\frac{d^2 x}{dt^2} = \frac{d^2 x_1}{dt^2} + \frac{d^2 x_2}{dt^2} = -k \left( \frac{1}{m_1} + \frac{1}{m_2} \right) x = -k \left( \frac{m_1 + m_2}{m_1 m_2} \right) x$This is the equation of simple harmonic motion $\frac{d^2 x}{dt^2} = -\omega^2 x$,where $\omega^2 = k \left( \frac{m_1 + m_2}{m_1 m_2} \right) = \frac{k}{\mu}$,where $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass.The time period $T$ is given by:$T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{m_1 m_2}{k(m_1 + m_2)}}$

Two masses $m_1$ and $m_2$ connected by a spring of spring constant $k$ rest on a frictionless surface. If the masses are pulled apart and let go,the time period of oscillation is

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