Question

In the arrangement shown in figure, pulleys are light and springs are ideal with equal spring constant of $100\text{N/m}$. The time period of small vertical oscillations of block of mass $1\text{kg}$ is

• A. $0.2\pi \text{s}$
• B. $0.4\pi \text{s}$
• C. $0.8\pi \text{s}$
• D. $0.6\pi \text{s}$

Answer 1

The correct option is C $0.8\pi \text{s}$

$\text{FBD}$ of the sytem:


If block will be displaced down by $x$ then each spring will extend by restoring force developed.

Let us suppose, restoring force is $F$, then extension in each spring will be $\frac{2F}{K}$.

So, the sum of extensions will be equal to the displacement of block $x$.

$2\left(\frac{2F}{K}\right)+2\left(\frac{2F}{K}\right)+2\left(\frac{2F}{K}\right)+2\left(\frac{2F}{K}\right)=x$

$\Rightarrow \frac{16F}{K}=x$

$\Rightarrow F=\frac{Kx}{16}$

Restoring force in the spring is equal to the force acting on the block.

$\Rightarrow ma=-\frac{Kx}{16}$
where, $a$ is acceleration of the block in downward direction.

Substituting the given values,

$\Rightarrow 1\times a=-\frac{100\times x}{16}$

$\Rightarrow \frac{x}{a}=-\frac{4}{25}$

Now, time period,

$T=2\pi \sqrt{\left|\frac{x}{a}\right|}$

$\Rightarrow T=2\pi \sqrt{\frac{4}{25}}$

$\Rightarrow T=\frac{4\pi }{5}=0.8\pi \text{s}$

Hence, option (c) is correct answer.