Question

A block of mass $4kg$ attached with spring of spring constant $100N/m$ is executing SHM of amplitude $0.1m$ on smooth horizontal surface as shown in figure. If another block of mass $5kg$ is gently placed on it, at the instant it passes through the mean position and new amplitude of motion is $n^{-1}$ meter then find $n$. Assuming that two blocks always move together.

Answer 1

The angular frequency where mass $m_2$ is $\omega =\sqrt{\frac{k}{m_1}}=\sqrt{\frac{100}{4}}=5rad/s$

When the mass is at mean position velocity, $v_0=\omega A$

$A$= amplitude before the mass $m_2$ is placed.$=0.1m$

$\Rightarrow v_0=5\times 0.1=0.5m$

Jst after the mass $m_2$ is placed at the time when $m_1$ passes mean position, the velocity of compound mass be, $v$. Applying conservation of linear momentm, we get, $m_1{v}_0=(m_1+m_2)v$

$\Rightarrow v=\frac{m_1}{m_1+m_2}{v}_0=\frac{4}{4+5}\times 0.5$

$\Rightarrow v=\frac{2}{9}m/s$

The new kietic energy of the system $\frac{1}{2}(m_1+m_2)v^2$

$=\frac{1}{2}9\times (\frac{2}{9}{)}^2=\frac{2}{9}J$

If the spring compressed by $x$ distance by this much kinetic energy, $x$ will be the new amplitude of motion. $\frac{1}{2}k{x}^2=\frac{2}{9}$

$\Rightarrow x^2=\frac{4}{9k}=\frac{4}{900}$

$\Rightarrow x=\frac{2}{30}$

$\Rightarrow x=\frac{1}{15}={15}^{-1}$

So, $n=15$