Answer to Question #132829 in Mechanics | Relativity for viccy

Question

a block of mass 2m is connected to a vertical massless spring and another block of mass m kept on 2m.find the amplitude of oscillation that mass m will not left mass 2m.

Answer 1

Explanations & Calculations

• Refer to the arrangement above & positive measuring direction is positive downward.
• If needed take the spring constant as k.
• Consider the system is released from a distance x down from the normal length level of the spring (then moving upward) and apply Newton's second law on the m kg block.

$\qquad\qquad \begin{aligned} \small R-mg&= \small m(-\ddot{x})\\ \small R&= \small m(g-\ddot{x}) > 0\\ \small \ddot{x}&<g \end{aligned}$ : R > 0 while in contact

• Then apply the same to the whole system to analyze the amplitude.

$\qquad\qquad \begin{aligned} \small F-3mg&= \small 3m(-\ddot{x})\\ \small- \ddot{x}&= \small \frac{kx-3mg}{3m}<g\\ \small x&>0 \end{aligned}$

• This means that block m stays on 2m in contact, only for the x values measured downward the spring's normal length/ only when the spring is compressed. As soon as the spring relaxes back to it's normal length, m block starts to move under gravity where R = 0.
• And this phenomenon is independent from the amplitude .

Amplitude = $\begin{aligned} \small \Big(x-\frac{3mg}{k}\Big) \end{aligned}$ where $\begin{aligned} \small \ddot{x} &= \small \frac{-k}{3m}\Big(x-\frac{3mg}{k}\Big) \end{aligned}$