The table below shows experimental data from a study on the vertical motion of a spring. Different m

Question

The table below shows experimental data from a study on the vertical motion of a spring. Different masses were attached to the spring, and the number of oscillations per second (i.e., the frequency) was measured.

Mass, $m$ kg	$0.01$	$0.02$	$0.04$	$0.06$	$0.08$
Frequency of oscillations, $f$	$20$	$14$	$10$	$8$	$7$

It is known that the mass $m$ (in kilograms) and the frequency $f$ are related by the formula:
$f^2 = \frac{1}{gm} \text{ where}\hspace{0.5em} g \text{ is a constant.}$

Plot a graph of $f^2$ against $\frac{1}{m}$ using the data in the table and draw a straight line of best fit.
Using your graph, estimate the following:
1. The frequency of oscillations when a mass of $0.07$ kg is attached to the spring.
2. The mass, to the nearest gram, that results in $15$ oscillations per second.
3. The value of the constant $g$.

The spring is then replaced with a different one. For this new spring, the relationship between $f$ and $m$ is given by: $f^2 = \frac{2}{m} + 100$.
On the same graph, draw the line that represents this new relationship.

Using your graph, find the mass that gives the same frequency of oscillations for both springs. Also, state the common frequency for this mass.

Accepted Answer

(a) $f^2\approx4\left(\frac1m\right)$. (b)(i) $7.56$ oscillations per second. (b)(ii) $18$ g. (b)(iii) $g\approx0.25$. (c) Approximately $20$ g and $14.1$ oscillations per second, using the displayed best-fit line.

Linear Law - Predictions from Linearised Models

Linear Law - Predictions from Linearised Models

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