91Ó°ÊÓ

In simple harmonic motion, the period of an oscillator is given by

$T=2\mathrm{Ï€}\sqrt{\frac{m}{k}}$

Note that $T$does not depend on the amplitude but only on the mass $m$and the force constant $k$.

To determine the spring constant, we must utilise the best-fit line of an acceptable graph.

Equation left is used to calculate the oscillator's period:

$T=2\mathrm{Ï€}\sqrt{\frac{m}{k}}$

Squaring both sides we obtain

$T^2=4{\mathrm{Ï€}}^2\frac{m}{k}$

$T^2=\left(\frac{4{\mathrm{Ï€}}^2}{k}\right)m$

A graph of $T^2$vs $m$ is a straight line starting at the origin with a slope equal to , according to this equation.

Make a new table for $T^2$and the associated $m$, remembering that the times in the given table are for 10 oscillations, therefore we must divide them by$10$in the new table.

Using graphing software, plot $T^2$versus $m$and calculate the equation of the best-fit line:

$T^2=\left(6.1253{\mathrm{s}}^2/\mathrm{kg}\right)m$

Comparing Equations (1) and (2), we obtain:

$\frac{4{\mathrm{Ï€}}^2}{k}=6.1253{\mathrm{s}}^2/\mathrm{kg}$

Solve for $k$

localid="1650122061136" $\begin{array}{r}k=\frac{4{\mathrm{Ï€}}^2}{6.1253{\mathrm{s}}^2/\mathrm{kg}}\\ =6.4\mathrm{N}/\mathrm{m}\end{array}$