Question

Figure shows a smooth horizontal table in $‘x-y^{\prime }$ plane between two identical fixed walls. Two springs are connected to the small ball. The length of the springs in the free state is $l$. The ball is shifted slightly from the equilibrium position in two different ways, once along the axis $OX$ and second along the y-axis and it begins to perform vibrations. The time period for these motions is (Assume vibrations to be small)

• A. motion along x-axis is simple harmonic
• B. motion along y-axis is simple harmonic
• C. motion along y-axis is not simple harmonic
• D. $T_x=2\pi \sqrt{\frac{m}{2k}}$

Answer 1

Answer: (A), (C), (D)

$T_x=2\pi \sqrt{\frac{m}{2k}}$

If we displace the ball along $x$ axis, then force along the line of motion will be sum of forces due to spring 1 pulling force and spring 2 pushing force as shown in the diagram below.

$F_{restoring}=-2kx$
$-m{\omega }^{2}x=-2kx$
$\omega =\sqrt{\frac{2k}{m}}$
$T_x=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{2k}}$

Now if displace the ball slightly along y axis, then total force on ball will be pulling forces by the springs.


$F_{restoring}=-2k{y}^{\prime }\sin \theta$
$F_{restoring}==-2k(\sqrt{l^2+y^2}-l)\frac{y}{\sqrt{l^2+y^2}}$
$F_{restoring}=-2k(1-\frac{l}{\sqrt{l^2+y^2}})y$
$F_{restoring}=-2k(1-\cos \theta )y$
$F_{restoring}=-2k\times 2{\sin }^2\frac{\theta }{2}\times y$

Now since $y$ is very small, so $\theta$ is very small as well, so we can write $2{\sin }^2\frac{\theta }{2}$ as $(\frac{\theta }{2}{)}^2$.
$\therefore F_{restoring}=-2k\times 2\times \frac{{\theta }^2}{4}\times y$
$\therefore F_{restoring}=-k{\theta }^{2}y$
Also $\theta =\tan \theta =\frac{y}{l}$
$\therefore F_{restoring}=-\frac{k{y}^3}{l^2}$
Clearly it is not first order harmonic motion. Hence it will not perform SHM along y axis.