Question

A disc of mass m is connected to two springs having spring constants $k_1$ and $k_2$ as shown in the figure. Find the time period of oscillation.

• A. $2\pi \sqrt{\frac{2m}{(k_1+4k_2)}}$
• B. $2\pi \sqrt{\frac{m}{(k_1+4k_2)}}$
• C. $2\pi \sqrt{\frac{3m}{2(k_1+4k_2)}}$
• D. $2\pi \sqrt{\frac{2m}{3(k_1+2k_2)}}$

Answer 1

The correct option is C $2\pi \sqrt{\frac{3m}{2(k_1+4k_2)}}$

At any instant centre of disc is displaced by x (towards left).
Then spring attached at C is compressed by x and spring attached at A elongates by 2x. Let v be the velocity of centre of cylinder and $\omega$ its angular velocity. Total mechanical energy in displaced position is
$E=\frac{1}{2}m{v}^2+\frac{1}{2}{I}_C{\omega }^2+\frac{1}{2}{k}_1{x}^2+\frac{1}{2}{k}_2(2x{)}^2$
But $\omega =\frac{v}{R}and{I}_C=\frac{1}{2}m{R}^2$
Hence $E=\frac{3}{4}m{v}^2+\frac{1}{2}{k}_1{x}^2+2k_2{x}^2$
In case of pure rolling energy is conserved
$\therefore \frac{dE}{dt}=0$

or, $\frac{3}{2}mv\left(\frac{dv}{dt}\right)+k_{1}x\left(\frac{dx}{dt}\right)+4k_{2}x\left(\frac{dx}{dt}\right)=0$
$\frac{dx}{dt}=v$ and $\frac{dv}{dt}=a\left(acceleration\right)$, with these substitution we get,
$\frac{3}{2}ma=-(k_1+4k_2)xi.e.,a\infty -x$
$T=2\pi \sqrt{\left|\frac{x}{a}\right|}=2\pi \sqrt{\left|\frac{3m}{2(k_1+4k_2)}\right|}$