Question

Calculate the angular frequency of the system shown in figure. Friction is absent everywhere and the threads, spring and pulleys are massless. Given that $m_A$ = $m_B$ = m.

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

Answer 1

Answer: (B)

Let $x_0$ be the extension in the spring in equilibrium.
Then equilibrium of $A$ and $B$ give,
$T=k{x}_0+mg\sin \theta \dots$ (i)
and $2T=mg\dots$ (ii)
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,
$E=\frac{1}{2}k{\left(x+x_0\right)}^2+\frac{1}{2}(m{)}_A(v{)}^2+\frac{1}{2}(m{)}_B{\left(\frac{v}{2}\right)}^2$
$+(m{)}_A(gh{)}_A-(m{)}_B(gh{)}_B$
or $E=\frac{1}{2}k{\left(x+x_0\right)}^2+\frac{1}{2}m(v{)}^2+\frac{1}{8}(mv{)}^2$
$+mgx\sin \theta -mg\frac{x}{2}$
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 $k{x}_2+mg\sin \theta =\frac{mg}{2}$ [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,
$T=2\pi \sqrt{\left|\frac{x}{a}\right|}=2\pi \sqrt{\frac{5m}{4k}}$
$\omega =\frac{2\pi }{T}=\sqrt{\frac{4k}{5m}}$