# Find value of k in SHM

2 votes
1,975 views

A block of mass 0.9 kg attached to a spring of force constant k is lying on a frictionless floor. The spring is compressed to $\sqrt2$ cm and the block is at a distance of 1/$\sqrt2$ 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 approximate value of k. My try asked Dec 27, 2016

## 1 Answer

2 votes

Best answer

Your answer is correct until the arithmetical error in the last line. $k=10\pi^2 N/m$, not $1000\pi^2$.

The mass performs SHM which is 'frustrated' by the wall. It does not perform complete oscillations. We can assume that the collision is "instantaneous" - which means that the collision time is much less than the period $T=0.2s$.

The amplitude of free oscillations (if the wall were not there) is $\sqrt2$, the maximum compression of the spring. The mass rebounds from the wall before it reaches the equilibrium position, since $\frac{1}{\sqrt2} \lt \sqrt2$.

The angular frequency of free oscillations is $\omega=\sqrt{\frac{k}{m}}$. Until it collides with the wall the equation of motion is $x=\sqrt2 \cos(\omega t)$ where $x$ is measured from the equilibrium position. The time $t \approx \frac12T=0.1s$ after maximum compression until the mass collides with the wall is given by
$\sqrt2-\frac{1}{\sqrt2}=\sqrt2 \cos(0.1\omega)$
$2-1=1=2\cos(0.1\omega)$
$\cos(0.1\omega)=\frac12=\cos(\frac13\pi)$
$\omega=\frac{10}{3}\pi$
$\omega^2=\frac{k}{m}=\frac{100}{9}\pi^2$
$k=\frac{100}{9}\pi^2 m = 10\pi^2 \approx 100N/m$.

answered Dec 27, 2016 by (28,746 points)
selected Dec 28, 2016 by koolman