Problem — Given the graph, find k

Problem

The oscillation period of a mass attached to a horizontal spring is measured, varying the mass. Determine the spring constant $k$, using the relation $T^2 = 4\pi^2 m/k$.

$m$ (kg)	0,25	1,0	2,25	4,0
$T$ (s)	0,50	1,00	1,50	2,00

Solution

From the relation $T^2 = 4\pi^2 m/k$ the ratio $T^2/m$ must be constant and equal to the slope of the line $T^2$ against $m$. Check on the data: $\begin{aligned} 0{,}50^2/0{,}25 = 1{,}00 \\ 1{,}00^2/1{,}0 = 1{,}00 \\ 1{,}50^2/2{,}25 = 1{,}00 \\ 2{,}00^2/4{,}0 = 1{,}00 \end{aligned}$ The slope is therefore $T^2/m = 1{,}0\;\text{s}^2/\text{kg}$. From $T^2/m = 4\pi^2/k$ I obtain $k = \frac{4\pi^2}{T^2/m} = \frac{4\pi^2}{1{,}0} \quad\Rightarrow\quad \ev{k \approx 39{,}5\;\text{N/m}}$

Links

Topics: Oscillazioni e moto armonico Concepts: Moto armonico semplice Skills: Lettura dei grafici · Analisi dimensionale Objects: Molla