Figure 4 shows the structure of a violin and Figure 5 shows a close-up image of the tuning pegs - AQA - A-Level Physics - Question 3 - 2018 - Paper 1

To show that the mass of a 1.0 m length of the string is about 4 × 10^{-4} kg, we can use the relationship between frequency, tension, and linear mass density:

$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

Where:

• $f$ is the frequency (370 Hz),
• $L$ is the vibrating length (0.33 m),
• $T$ is the tension (25 N), and
• $\\mu$ is the linear mass density of the string.

Rearranging gives us:

$\mu = \frac{T}{(2Lf)^2}$

Substituting the known values:

$\mu = \frac{25}{(2 \times 0.33 \times 370)^2} ≈ 4 × 10^{-4} \text{ kg/m}$

Thus, for a 1.0 m length, the mass would approximately be 4 × 10^{-4} kg.

The speed of the waves in the string can be determined using the formula:

$v = f \times \lambda$

Where:

• $f$ is the frequency (370 Hz), and
• $\lambda$ is the wavelength.

In the case of the first harmonic, the wavelength can be calculated as:

$\lambda = 2L = 2 \times 0.33 = 0.66 \, m$

Thus, substituting:

$v = 370 \times 0.66 ≈ 244.2 \, m/s$

From Figure 6, the tension in the string is directly proportional to the extension. When the initial tension is 25 N, and based on the graph, we can determine the tension for different extensions. If the tension increases steadily with extension as shown, the tension will eventually become greater than the original tension as the string is stretched. Therefore, the relationship can be observed through linearity and should remain consistent up to the breaking point.

To find the new frequency after rotating the tuning peg, we first calculate the new length of the string assuming the angle of rotation causes an increase in the effective length. The increase in string length is found through:

$\text{extra extension} = 22.5 \times \frac{75}{360} \approx 3.75 \text{ mm}$

Then,

• New length $L_{new} = 0.33 + (3.75 / 1000) = 0.33475 \, m$.

Now using the formula: $f_{new} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ Using the updated parameters, we can compute the new frequency, which will be greater than the original (greater than 370 Hz). Thus the final value should be determined by calculation.