Question

If the mass of the pulleys shown in figure is very very small and the cord is inextensible, the angular frequency of oscillation of the system is

• A. $\sqrt{\frac{k_a+k_b}{m}}$
• B. $\sqrt{\frac{k_a{k}_b}{(k_a+k_b)m}}$
• C. $\sqrt{\frac{k_a{k}_b}{4m(k_a+k_b)}}$
• D. $\sqrt{\frac{4k_a{k}_b}{(k_a+k_b)m}}$

Answer 1

The correct option is C $\sqrt{\frac{k_a{k}_b}{4m(k_a+k_b)}}$

Let $T$ be the tension in the cord and $x_a$ and $x_b$ the displacement of pulleys $A$ and $B$ respectively.
Now assume that pulley $A$ is fixed; the extension in spring will be $x_b=\frac{x}{2}$, as the block gets lowered by $x$
$\Rightarrow x=2x_b$
Similarly if we imagine that pulley $B$ is fixed, $x=2x_a$.

However neither pulley $B$ nor pulley $A$ is fixed.
$\Rightarrow x=2x_a+2x_b...\left(1\right)$
From balancing force on pulleys( pulley is massless)


$2T=k_b{x}_b...\left(2\right)$
and $2T=k_a{x}_a...\left(3\right)$
If $k_{eq}$ denotes equivalent spring constant, for the spring block system.
$\Rightarrow T=k_{eq}x$
$\Rightarrow \frac{T}{k_{eq}}=x$
$\Rightarrow \frac{T}{k_{eq}}=2x_a+2x_b....\left(4\right)$
From equations $\left(2\right)$ and $\left(3\right)$, $\left(4\right)$
$x_a=\frac{2T}{k_a}$ and $x_b=\frac{2T}{k_b}$
and $\Rightarrow k_{eq}=\frac{1}{4(\frac{1}{k_a}+\frac{1}{k_b})}$
Hence angular frequency of oscillation is,
$\omega =\sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{k_a{k}_b}{4m(k_a+k_b)}}$