Question

Two blocks $A\left(2kg\right)$ and $B\left(3kg\right)$ rest up on a smooth horizontal surface are connected by a spring of stiffness $120$ N/m. Initially the spring is un-deformed. $A$ is imparted a velocity of $2$ m/s along the line of the spring away from B. Find the displacement of $A$, $t$ seconds later.

• A. $1.6t+0.24sin10t$
• B. $0.8t+0.12sin10t$
• C. $1.6t+0.12sin10t$
• D. $Noneofthese$

Answer 1

The correct option is B $0.8t+0.12sin10t$

From fig, we can say an extension of spring $(x_2-x_1)$

A reduced mess of the system $\mu =\frac{m_A{\times m}_B}{m_A+m_B}=\frac{3\times 2}{3+2}=1.2kg$

Angular frequency of oscillation

$\omega =\sqrt{\frac{k}{\mu }}=\sqrt{\frac{120}{12}=10}rad|s$

$\therefore$ displacement

$x=x_0\sin \left(10t\right)$ here $x=x_A-x_B$

Here $x_A-x_B=x_0\sin \left(10t\right)...\left(1\right)$

The velocity of C.O.M of the system

$v_{cm}=\frac{m_A.v_A+m_B.v_B}{m_A+m_B}=\frac{2\times 2+3\times 0}{2+3}=0.8m/s=constant$

$\therefore$ Displacement of c.m. in time 't' is

$x_{cm}=v_{cmt}\Rightarrow x_{cm}=0.8t$

But $x_{cm}=\frac{2x_A+3x_B}{2+3}\therefore \frac{2x_A+3x_B}{5}=0.8t\Rightarrow {2x}_A+3x_B=4t\to \left(iii\right)$

Solving (1)(2) we get

$5x_A=3x_0\sin \left(10t\right)+4t\Rightarrow x_A=0.6_0{x}_0\sin \left(10t\right)+0.8t$

Now we need to find out $x_0$(max.extention od spring)

At point of max .extention the velocity of two block is $v_{cm}=0.8m/s$

Fromenergy conservation

$\frac{1}{2}\times m_A\times v_A^2=\frac{1}{2}\times x\times x_0^2+\frac{1}{2}\times m_A\times v_{cm}^2+\frac{1}{2}{m}_B\times v_{cm}^2\Rightarrow \frac{1}{2}\times 2\times 2^2=\frac{1}{2}\times 120\times x_0^2+\frac{1}{2}\times 2\times 0.8^2+\frac{1}{2}\times 3\times 0.8^2\Rightarrow 60x_0^2=4^2-0.69-0.96\Rightarrow x_0^2=\frac{2.4}{60}=0.04\Rightarrow x_0=0.2m{x}_A=0.2\sin \left(10t\right)+0.8t$