Question

One end of an ideal spring is fixed to a wall at origin $O$ and axis of spring is parallel to x-axis. A block of mass $m=1kg$ is attached to free end of the spring and it is performing SHM. Equation of position of the block in coordinate system shown in figure is $x=10+3\sin \left(10t\right)$. Here, $t$ is in second and $x$ in cm. Another block of mass $M=3kg$, moving towards the origin with velocity $30cm/s$ collides with the block performing SHM at $t=0$ and gets stuck to it. Calculate new amplitude of oscillations.

Answer 1

${\omega }^2=\frac{k}{m}$
$\therefore$ $k=m{\omega }^2=\left(1\right){\left(10\right)}^2=100$ N/m
At $t=0$, block of mass m is at mean position (x=10 cm) and moving towards positive x-direction with velocity $A\omega$ or $30cm/s$.
From conservation of linear momentum,
$(M+m)v=M\left(30\right)-m\left(30\right)$
Substituting the values, we have
$v=15cm/s$ or $0.15m/s$
$v=15cm/s$
From conservation of mechanical energy,
$\frac{1}{2}(M+m)v^2=\frac{1}{2}k{A}^2$
$\therefore$ $A=\left(\sqrt{\frac{M+m}{k}}\right)v={\left(\frac{4}{100}\right)}^{1/2}\left(0.15\right)$
$=0.03$ m
or $A=3$ cm