Question

A block of mass $100\text{gram}$ attached to a spring of stiffness $25\text{N/m}$ is lying on a frictionless floor as shown in figure. The block is moved to compress the spring by $8\text{cm}$ and released. If collision with the wall is elastic, find the time period of oscillations.

• A. $0.451\text{sec}$
• B. $0.5\text{sec}$
• C. $0.265\text{sec}$
• D. $1\text{sec}$

Answer 1

The correct option is C $0.265\text{sec}$


Given,
Mass of block $\left(\right(m)=100\text{g}$
Spring constant $\left(k\right)=25\text{N/m}$
Compression in the spring $\left(A\right)=8\text{cm}$
Separation between wall and block $\left(x\right)=4\text{cm}$
Time period $\left(T\right)=2\pi \sqrt{\frac{m}{k}}$
$=2\times 3.14\times \sqrt{\frac{0.1}{25}}$
$=0.397\text{s}$
From, $x=A\sin \omega t$
time taken for particle to reach from mean position to the wall is
$4=8\sin \omega t$
$\Rightarrow \sin \omega t=\frac{1}{2}$ (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
$\frac{T}{2}+2t$
$\Rightarrow \frac{T}{2}+\frac{T}{6}=\frac{2T}{3}=\frac{2}{3}\times 0.397=0.265\text{s}$