Question

One end of each of two identical springs, each of force-constant 0.5 N/m attached on the opposite sides of a wooden block of mass 0.01 kg. The other ends of the springs are connected to separate rigid supports such that the springs are unstretched and are collinear in a horizontal plane. To the wooden piece is fixed a pointer which touches a vertically moving plane paper. The wooden piece, kept on a smooth horizontal table is now displaced by 0.02 m along the line of springs and released. If the speed of paper is 0.1 m/s, find the equation of the path traced by the pointer on the paper and the distance two consecutive maximas on this path.

Answer 1

On the vertically moving paper the pointer traces out the path which gives us the equation of displacement of mass at any time $t$ . i.e, displacement as a function of $t$.

The two spring will apply some force on mass $m$ and it will add up. So,

$F=-(k+k)x$

$\Rightarrow F=-2kx$

$\Rightarrow m\frac{d^{2}x}{d{t}^2}=-2kx$

$\Rightarrow \frac{d^{2}x}{d{t}^2}+\left(\frac{2k}{m}\right)x=0$

$\Rightarrow \frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$

$\frac{d^{2}x}{d{t}^2}=$ Acceleration of mass $m$.

$\omega =\sqrt{\frac{2k}{m}}=\frac{2\times 0.5}{0.01}$

$=\sqrt{100}=10rad/s$

The general solution of this equation gives the equation of path traces out by pointer.

So,

$x=A\sin \omega t$ (In general solution)

$\Rightarrow x=0.02\sin 10t$ ($A$= Maximum displacement from mean position $=0.02m$ $2\times {10}^{-2}m$)

Now velocity of paper is $0.1m/s$. And two consecutive maximum for $x$ occurs at time:

$t_1=\frac{\pi }{20}sec$ and

$t_2=\frac{21\pi }{20}sec$

The time interval between two consecutive maxima:

$△t=t_2-t_1=\pi sec$

In this time paper will be moved by: $u△t=0.1\times \pi =0.1\pi m$

in vertical direction, So it should be the distance between two consecutive maxima in the paper.