Question

Two masses ${m_1}$and ${m_2}$ are suspended together by a massless spring of constant K. When the masses are in equilibrium, ${m_1}$is removed without disturbing the system. The amplitude of oscillations is

A. $\dfrac{{{m_1}g}}{K}$
B. $\dfrac{{{m_2}g}}{K}$
C. $\dfrac{{({m_1} + {m_2})g}}{K}$
D. $\dfrac{{({m_1} - {m_2})g}}{K}$

Answer 1

Hint:The problem is from the oscillations and waves section of physics. We have to apply the concepts of spring constant and vibrations to solve this problem. We can write the equation of motion for the two cases. When two masses are attached and when one mass is removed. After that from those two equations, we can find the amplitude of the oscillations.

Complete step by step solution:
The diagram is redrawn as

The length of the initial spring is $l$. When mass ${m_2}$ is attached to the spring and it stretches a distance x from point B to C. When mass \[{m_1}\] is attached to the spring and it stretches a distance y from point C to D. Then the equation of motion will be,
$({m_1} + {m_2})g = - k(x + y)$...........(1)
When mass ${m_1}$ is removed from the system, the equation of motion will become
${m_2}g = - kx$ …………(2)
By comparing and solving equations (1) and (2) we will get,
${m_1}g = - ky$
$\therefore y = - \dfrac{{{m_1}g}}{k} = \left| {\dfrac{{{m_1}g}}{k}} \right|$
That is, the amplitude of the oscillation is $y = \dfrac{{{m_1}g}}{k}$.

Hence, the correct option is option A.

Additional Information: The process of any quantity or measure fluctuating repeatedly about its equilibrium value in time is known as oscillation. A periodic change in an object's value between two values or around its central value is another way to define oscillation.

Note: The spring constant is defined as the stiffness of the spring. The equation of the spring constant is given as, \[k = - \dfrac{F}{x}\]. Where F = Force applied, x = displacement by the spring and k = spring constant.