Question

A block of mass $100\text{g}$ attached to a spring of spring constant $100{\text{Nm}}^{-1}$ is lying on a frictionless floor as shown. The block is moved to compress the spring by $10\text{cm}$ and then released. If the collisions with the wall in front are elastiac, then the time period of the motion is

• A. $0.2\text{s}$
• B. $0.1\text{s}$
• C. $0.5\text{s}$
• D. $0.132\text{s}$

Answer 1

The correct option is D $0.132\text{s}$

Given,
Mass of block $\left(\right(m)=100\text{g}$
Spring constant $\left(K\right)=100\text{N/m}$
Mass compression $\left(A\right)=10\text{cm}$
Separation between wall and block $\left(x\right)=5\text{cm}$
Time period $\left(T\right)=2\pi \sqrt{\frac{m}{K}}$
$=2\times 3.14\times \sqrt{\frac{0.1}{100}}$
$=0.198\text{s}$
From, $x=A\sin \omega t$
time taken for particle to reach from mean position to the wall is
$5=10\sin \omega t$
$\Rightarrow \sin \omega t=\frac{1}{2}$ (or) $\omega t=\frac{\pi }{6}$
$\Rightarrow t=T/12$
Now time of oscillation is half of the time period + timetaken for particle to reach the wall from mean position and from wall to mean position
$\frac{T}{2}+2t$
$\Rightarrow \frac{T}{2}+\frac{T}{6}=\frac{2T}{3}$
$=\frac{2}{3}\times 0.198$
$=0.132\text{s}$