Question

Rod is massless and of length $\ell$. Find angular frequency $\omega$ (in $s ^{-1}$ ) of small oscillation of the block of mass $m$, the hinge is frictionless. $k _{1}, k _{2}, k _{3}$ are spring constants of the shown springs. Ignore gravity. ( $\left.k _{1}= k _{2}=10\, N / m , k _{3}=\frac{15}{4} N / m , m =250\, gm \right)$

Answer 1

$F =- k _{1} x - k _{2}\left( x -\frac{\ell}{2} \theta\right)$
$k _{2}\left( x -\frac{\ell}{2} \theta\right) \frac{\ell}{2}= k _{3} \ell \theta \ell$
$k _{2}\left( x -\frac{\ell}{2} \theta\right)=2 k _{3} \ell \theta$
$\left(\frac{ k _{2}}{2}+2 k _{3}\right) \ell \theta= k _{2} x$
$F =- k _{1} x - k _{2} x +\frac{ k _{2}}{2} \ell \theta$
$F=-k_{1} x-k_{2} x+\frac{k_{2}}{2}\left(\frac{k_{2} x}{\frac{k_{2}}{2}+2 k_{3}}\right)$
$F=-10 x-10 x+5\left(\frac{10 x}{5+\frac{15}{2}}\right)$
$F=-20 x+4 x$
$F=-16\, x$
$\Rightarrow a=-\frac{16}{m} x$
$\omega=\sqrt{\frac{16}{m}}$
$\omega=\sqrt{\frac{16}{\frac{1}{4}}}$
$\omega=8\, rad / s$