A student investigated the variation of $f$, the fundamental frequency of a stretched string, with its length $l$ - Leaving Cert Physics - Question 2 - 2016

1. Plot the frequency $f$ on the y-axis (in Hz) and the length $l$ on the x-axis (in cm).
2. Label the axes clearly: 'Frequency (Hz)' and 'Length (cm)'.
3. Mark the recorded data points from the table on the graph.
4. Draw a straight line of best fit through the plotted points to show the relationship between $f$ and $l$. The graph should ideally show an inverse relationship, indicating that as length $l$ decreases, the frequency $f$ increases.

The relationship is given by the equation $f \propto \frac{1}{l}$, meaning that the frequency of the string is inversely proportional to its length.

This indicates that as the length of the string decreases, the frequency increases. The straight line through the origin on the graph supports this by showing that there is a linear inverse relationship between frequency and length.

To find the length of the string at a frequency of 192 Hz:

1. Locate the corresponding frequency of 192 Hz on the graph.
2. Draw a horizontal line from 192 Hz until it intersects the line of best fit.
3. From this intersection, draw a vertical line down to the x-axis to read the corresponding length $l$.
4. Based on the graph, the length $l$ is approximately 1.52 m$^{-1}$.

From the equation $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where:

• $f$ is the frequency,
• $L$ is the length of the string,
• $T$ is the tension, and
• $\mu$ is the mass per unit length.

Rearranging gives: $\mu = \frac{T}{(2f)^2}$

1. Use the read value of $f$ from the graph for a specific $l$.
2. Insert the tension $T = 8.5 N$ and computed $l$ to obtain $\\mu$.
3. The calculated mass per unit length of string is approximately $\mu \approx 1.3 \times 10^{-4} kg m^{-1}$.