Discussion Forum

A particle of mass m is attached to four springs with spring constant k , k,2k and 2k as shown in the figure. Four springs are attached to the four corners of a square and a particle is placed  at the centre . If the particle is pushed slightly  towards  any sides of the square and released, the period of oscillation will be 

Answer:

Explanation:

 along Y-axis , net force on the particle is 

  $F=-(kx \sin 45^{0}+kx \sin 45^{0}+2kx \sin 135^{0}+2kx \sin 135^{0})$

  =    $-\frac{6}{\sqrt{2}}$kx

 $\Rightarrow$     $m\frac{d^{2}x}{dt^{2}}=-3\sqrt{2}kx\Rightarrow\frac{d^{2}x}{dt^{2}}+\frac{3\sqrt{2}kx}{m}=0$

  This is equation of SHM

 Angular frequency,   $\omega^{2}=3\sqrt{2}\frac{k}{m}$

$\Rightarrow$     $f=\left(\frac{1}{2}\pi\right)\sqrt{\frac{3\sqrt{2}k}{m}}\Rightarrow T=2\pi\sqrt{\frac{m}{3\sqrt{2}k}}$