Question

One end of each of the two identical springs, having force constant $0.5\text{N/m}$ are attached on the opposite ends of a wooden block of mass $0.01\text{kg}$. The other ends of the spring are attached 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 perpendicularly moving plane paper on horizontal plane.

The wooden piece kept on a smooth horizontal table is now displaced by $0.02\text{m}$ along the line of springs and released. If the speed of the paper perpendicular to the spring’s length is $0.1\text{m/s}$, find the equation of the path traced by the pointer on the paper and the distance between two consecutive maxima on this path.

• A. $y=0.02\text{sin}\left(100x\right)$, $\frac{\pi }{50}m$
• B. $y=0.02\text{sin}\left(50x\right)$, $\frac{\pi }{50}m$
• C. $y=0.04\text{sin}\left(100x\right)$, $\frac{\pi }{25}m$
• D. $y=0.04\text{sin}\left(50x\right)$, $\frac{\pi }{25}m$

Answer 1

The correct option is A $y=0.02\text{sin}\left(100x\right)$, $\frac{\pi }{50}m$



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 $t$.

$y=A\sin \left(kx\right)---\left(i\right)A=0.02mk=\frac{\omega }{v}$

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

$F_{net}=-2kx$
$\Rightarrow F_{net}=-2kx$
$\Rightarrow m\frac{d^{2}x}{d{t}^2}=-2kx$
$\Rightarrow \frac{d^{2}x}{d{t}^2}+\left(\frac{2k}{m}\right)x=0$
$\Rightarrow \frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$
$\frac{d^{2}x}{d{t}^2}=$ Acceleration of mass $m$
$\omega =\sqrt{\frac{2k}{m}}=\frac{2\times 0.5}{0.01}$
$=\sqrt{100}=10rad/s$

From (i)
$\begin{array}{cc}y& =A\sin kx\\ & =\left(0.02m\right)\sin 100x\end{array}$.
Now the distance between two consecutive maxima on this path is wavelength ($\lambda$).
So, $\lambda =\frac{2\pi }{k}=\frac{2\pi }{100}=\frac{\pi }{50}m$