Question

You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel to plot T2 (in s2) on the y-axis versus M (in kg) on the x-axis. The equation for the straight line that gives the best fit to your data is
y = 0.0569x + 0.00100.
Units are not specified but are determined by the units of what is being graphed.
(a) What is the force constant k of the spring?


(b) What is the mass of the spring?

Answer 1

We the Period T we can find the constant k,

That is

$T = 2 \pi \sqrt{\frac{m}{k}}$

We delete the squart root elevating squareing everything,

$T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}$

M is the hanging mass, m is the spring mass,

k is the spring constant and T the time period

a) So for the equation we can compare, that is,

$y=T^2=0.0569x+0.0010$

Here x is the hanging mass M, so comparing the equation we know that

$\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m$

b) For the mass of the spring we make similar process, so comparing,

$\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g$