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. Values of $f$ and $l$ are plotted: The values from the data table are plotted with $f$ on the y-axis and $l$ on the x-axis.

2. Axes labelled correctly: The y-axis should be labeled as 'Frequency (f in Hz)' and the x-axis should be labeled as 'Length (l in cm)'.

3. Points plotted accurately: Each data point corresponds to pairs of $(f, l)$ from the table.

4. A straight line with good fit: After plotting the points, draw a straight line that best fits the trend of the points leading to a downward sloping graph indicating an inverse relationship.

The relationship can be stated as:

$f \propto \frac{1}{\sqrt{l}}$

This implies that as the length of the string increases, the frequency decreases. The graph verifies this since it shows a negative slope, proving that longer strings resonate at lower frequencies.

(i) To determine the length of the string at a frequency of 192 Hz:

1. Length read from graph: After finding where 192 Hz intersects on the graph, it should show approximately $l_i = 1.52$ m.
2. Value inverted: The corresponding value of the length for this frequency is determined using the established relationship.

(ii) To calculate the mass per unit length, use the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

Given that $T = 8.5$ N and using the calculated values of $L$ and $f = 192$ Hz:

1. Substitute into the equation to find $\,\mu$:
   • Rearranging gives: $\mu = \frac{T}{(2Lf)^2}$
2. Values can then be plugged in to find the mass per unit length approximately as $\mu \approx 1.3 \times 10^{-4} \; kg \; m^{-1}$.