Question

A block of mass $0.9kg$ attached to a spring of force constant $k$ is lying on a frictionless floor. The spring is compressed to $\sqrt{2}cm$ and the block is at a distance $1/\sqrt{2}cm$ 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 $0.2sec$. Find the approximate value of $k$.

Answer 1

The given block will perform SHM because floor is frictionless. It does not perform complete oscillations because of the wall. $T=0.2sec$, Also the collision time is much less than time period $T$. Amplitude of free oscillations if wall was not there is $\sqrt{2}$, Which is maximum compression of the spring. the block will make elastic collision with wall because $\frac{1}{\sqrt{2}}<\sqrt{2}.$

For free oscillations, $\omega =\sqrt{\frac{k}{m}}$

$x=\sqrt{2}cos\omega t$

But, $t\approx \frac{1}{2}T=0.1sec$ After maximum compression until the mass collide with the wall eq is given as,

$\sqrt{2}-\frac{1}{\sqrt{2}}=\sqrt{2}cos\left(0.1\omega \right)$

$2-1=2cos\left(0.1\omega t\right)$

$cos\left(0.1\omega t\right)=\frac{1}{2}=cos\left(\frac{1}{3}\pi \right)$

$\omega =10\frac{\pi }{3}$

${\omega }^2=\frac{k}{m}=\frac{100}{9}{\pi }^2$

$k=\frac{100}{9}{\pi }^{2}m\approx 100N/m$