Question

A block of mass $0.9kg$ attached to a spring of force constant $k$ is lying on a friction less floor.The spring is compressed by $\sqrt{2}cm$ and the block is at a distance $\frac{1}{\sqrt{2}}cm$ from the wall as shown in the figure. When the block is released, it makes elastic collision with the wall and its period of motion is $0.2\sec$. Find the appropriate value of $k$ in $N{m}^{-1}$

Answer 1

$x=A\sin \left(\omega t\right)$
Let $x=\frac{1}{\sqrt{2}},A=\sqrt{2},\omega =\frac{2\pi }{T},T=0.2\sec$
Time required to reach wall is given by
$\frac{1}{\sqrt{2}}=\sqrt{2}\sin \left(\frac{2\pi t}{0.2}\right)$
$\sin \left(10\pi t\right)=\frac{1}{2}\Rightarrow t=\frac{1}{60}$
solving this we get
$t=T/6$
Now total time of motion is
$t=2\times T/6$
$t=T/3$
By problem $t=0.2$ sec
$T=0.6$ sec
$T=2\pi \times \sqrt{m/k}$
By solving we get $K=100N/m$