Question

Friction is absent everywhere and the threads, spring and pulleys are massless. If $m_A=m_B=m$ , then the angular frequency of the system for small oscillations will be

• A. $\sqrt{\frac{\text{2k}}{\text{4m}}}$
• B. $\sqrt{\frac{\text{4k}}{\text{5m}}}$
• C. $\sqrt{\frac{\text{6k}}{\text{7m}}}$
• D. $\sqrt{\frac{\text{8k}}{\text{5m}}}$

Answer 1

Answer: (B)

Let $x_0$ be the extension in the spring in equilibrium. Then
equilibrium of $A$ and $B$ given,
$T=k{x}_0+mg\sin \theta$
and $2T=mg$
Here, $T$ is the tension in the string. Now, suppose $A$ is further displaced by a distance $x$ from its mean position and $v$ be its speed at this moment. Then $B$ lowers by $\frac{x}{2}$ and speed of $B$ at this instant will be $\frac{v}{2}$ Total energy of the system in this position will be,
$E=\frac{1}{2}k{\left(x+x_0\right)}^2+\frac{1}{2}{m}_A{v}^2+\frac{1}{2}{m}_B{\left(\frac{v}{2}\right)}^2+m_{A}g{h}_A-m_{B}g{h}_B$
or $E=\frac{1}{2}k{\left(x+x_0\right)}^2+\frac{1}{2}m{v}^2+\frac{1}{8}m{v}^2+mgx\sin \theta -mg\frac{x}{2}$
or $E=\frac{1}{2}k{\left(x+x_0\right)}^2+\frac{5}{8}m{v}^2+mgx\sin \theta -mg\frac{x}{2}$
Since, $E$ is constant,
$\frac{dE}{dt}=0$
or $0=k\left(x+x_0\right)\frac{dx}{dt}+\frac{5}{4}mv\left(\frac{dv}{dt}\right)+mg(\sin \theta )\left(\frac{dx}{dt}\right)-\frac{mg}{2}\left(\frac{dx}{dt}\right)$
Substituting, $\frac{dx}{dt}=v$
$\frac{dv}{dt}=a$
and $k{x}_0+mg\sin \theta =\frac{mg}{2}$
[From equations (i) and (ii)]
We get, $\frac{5}{4}ma=-kx$
Since, $a\propto -x$
Motion is simple harmonic, time period of which is,
$\begin{array}{l}T=2\pi \sqrt{\left|\frac{x}{a}\right|}=2\pi \sqrt{\frac{5m}{4k}}\\ \omega =\frac{2\pi }{T}=\sqrt{\frac{4k}{5m}}\end{array}$