Question

Two masses $M$ and $m$ are suspended together by a massless spring of force constant $k$. When the masses are in equilibrium, $M$ is removed without disturbing the system. Then the amplitude of oscillation is :

• A. $Mg/k$
• B. $mg/k$
• C. $(M+m)g/k$
• D. $\left(Mm\right)g/k$

Answer 1

Answer: (A)

$Mg/k$

Let the initial length of the spring is $l$. When one mass $m$ is suspended, the string will stretch by amount $x$ from B to C and when mass $M$ is suspended, the string will stretch by amount $y$ from C to D.

Thus, the equation of motion of the system, $(M+m)g=-k(x+y)....\left(1\right)$

When M is removed, the equation of motion will be , $mg=-kx...\left(2\right)$

Using (1) and (2), we get $Mg=-ky$

or $y=-Mg/k$

Thus the amplitude of the oscillation is $y=Mg/k$

(where negative sign indicates that the displacement is directed towards the mean position )