Question

In the system shown in the figure, the string, springs and pulley are weightless. The force constants of the two springs are $k_1=k$ and $k_2=2k$. Block of mass $M$ is pulled vertically down from its equilibrium position and released. Calculate the angular frequency of oscillation.
[Assume the top surface of the block (represented by line $AB$) always remains horizontal]

• A. $\omega =\sqrt{\frac{4k}{3M}}$
• B. $\omega =\sqrt{\frac{8k}{3M}}$
• C. $\omega =\sqrt{\frac{2k}{3M}}$
• D. $\omega =\sqrt{\frac{k}{3M}}$

Answer 1

The correct option is B $\omega =\sqrt{\frac{8k}{3M}}$

Let $x_0$ be the elongation in spring of force constant $k_2$ at equilibrium.
From the data given in the question, it is clear that the elongation in the other spring must be $2x_0$, so that tension is the same in both the springs.
[since both the springs are joined by the same piece of string, spring forces in each must be equal to tension in the rope]
For equilibrium of block,
$2k{x}_0+k\left(2x_0\right)=Mg$
$\Rightarrow 4k{x}_0=Mg\dots \dots \left(1\right)$
Let the block be displaced downward by $x$, causing the two springs to stretch further by $x_1$ and $x_2$ respectively.
Again, $k_1{x}_1=k_2{x}_2\Rightarrow x_1=2x_2$
and $x_1+x_2=2x$ [from string constraint]
$\therefore x_2=\frac{2x}{3}$ and $x_1=\frac{4x}{3}$

For the block displaced by distance $x$ from equilibrium/mean position,
$M\frac{d^{2}x}{d{t}^2}=Mg-k_1(x_1+2x_0)-k_2(x_2+x_0)$
$\Rightarrow M\frac{d^{2}x}{d{t}^2}=Mg-k_1(x_1+2x_0)-k_2(x_2+x_0)$
$\Rightarrow M\frac{d^{2}x}{d{t}^2}=Mg-k{x}_1-2k{x}_2-(2k+2k)x_0$
$\Rightarrow M\frac{d^{2}x}{d{t}^2}=-\frac{4kx}{3}-\frac{4kx}{3}$ [using (i)]
$\Rightarrow \frac{d^{2}x}{d{t}^2}=-\frac{8k}{3M}x$
Comparing this with $a=-{\omega }^{2}x$,
we get, $\omega =\sqrt{\frac{8k}{3M}}$
Thus, option (b) is the correct answer.