In an experiment the tension of a vibration string was kept constant while its length was varied so as to tune the string to series of vibrating tuning forks. A student obtained the following result: Draw a graph for the set of observation and fit the "best" straight line to them.

Calculate the inverse of the length, $\frac{1}{L}$, for each of the first four data points:

• For $L = 78.1 \text{ cm}$, $\frac{1}{L} = \frac{1}{78.1} \approx 0.01280 \text{ cm}^{-1}$
• For $L = 69.8 \text{ cm}$, $\frac{1}{L} = \frac{1}{69.8} \approx 0.01433 \text{ cm}^{-1}$
• For $L = 62.5 \text{ cm}$, $\frac{1}{L} = \frac{1}{62.5} = 0.01600 \text{ cm}^{-1}$
• For $L = 52.1 \text{ cm}$, $\frac{1}{L} = \frac{1}{52.1} \approx 0.01919 \text{ cm}^{-1}$

Plot frequency $f$ (in Hz) on the y-axis and $\frac{1}{L}$ (in cm⁻¹) on the x-axis. Mark the four data points: $(0.01280, 246)$, $(0.01433, 288)$, $(0.01600, 320)$, and $(0.01919, 384)$. Draw a straight line that best fits these four points. This line should pass close to the origin, as theoretically, $f=0$ when $L \to \infty$ (i.e., $\frac{1}{L}=0$).

Let $x = \frac{1}{L}$ and $y = f$.
$n = 4$
$\sum x = 0.01280 + 0.01433 + 0.01600 + 0.01919 = 0.06232$
$\sum y = 246 + 288 + 320 + 384 = 1238$
$\sum x^2 = (0.01280)^2 + (0.01433)^2 + (0.01600)^2 + (0.01919)^2 = 0.000993445$
$\sum xy = (0.01280 \times 246) + (0.01433 \times 288) + (0.01600 \times 320) + (0.01919 \times 384) = 19.766$

The formulas for linear regression are:

Substitute the calculated sums:

Rounding to four significant figures, the slope is $\boxed{\text{21160 Hz}\cdot\text{cm}}$.

Rounding to three significant figures, the intercept is $\boxed{\text{-16.6 Hz}}$.

The theoretical relationship for a vibrating string is $f = k \cdot \frac{1}{L}$, where $k$ is a constant.

Using the calculated slope ($m$) and intercept ($c$) from part (b), the experimental relationship is in the form $f = m \left(\frac{1}{L}\right) + c$.
Substituting the values:

The relationship is $\boxed{f = \frac{21160}{L} - 16.6}$, where $f$ is in Hz and $L$ is in cm.

Given $L = 45.0 \text{ cm}$.

Rounding to one decimal place, the frequency is $\boxed{\text{453.6 Hz}}$.