Question

A bock of mass $0.9\text{kg}$ is attached to a spring of force constant k is lying on a frictionless floor. The spring is compressed to $\sqrt{2}\text{cm}$ and the block is at a distance of $\frac{1}{\sqrt{2}}\text{cm}$ from the wall as shown in the figure. When the block is released, it makes an elastic collision with the wall and its period of motion is $0.2\text{s}$. Find the approximate value of $k$.

• A. $400\text{N/m}$
• B. $200\text{N/m}$
• C. $150\text{N/m}$
• D. $300\text{N/m}$

Answer 1

The correct option is A $400\text{N/m}$

Since while compression, though block reaches its extreme position $\sqrt{2}cm$, but during extension block is not able to reach its other extreme position because of elastic collision with wall. So phasor diagram of the block will look somewhat like this:

Where $\varphi$ can be calculated as,
$\cos \varphi =\frac{x}{A}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$
$\Rightarrow \varphi =\frac{\pi }{3}$

Now for a standard spring block system for whole $2\pi$ period,
$T=2\pi \sqrt{\frac{m}{k}}$

For the system in consideration in this question, total time taken will be for $(2\pi -\varphi )=\frac{4\pi }{3}$ phase.
So time period corresponding to $\frac{4\pi }{3}$ phase
$t=\frac{4\pi }{3}\sqrt{\frac{m}{k}}=0.2s$
Substituting values we get,
$k=400N/m$