Figure 1 shows an arrangement used by a student to investigate vibrations in a stretched nylon string of fixed length l. He measures how the frequency f of first-harmonic vibrations for the string varies with the mass m suspended from it. Table 1 shows the results of the experiment. Table 1 | m / kg | f / Hz | |--------|--------| | 0.50 | 110 | | 0.80 | 140 | | 1.20 | 170 | Show that the data in Table 1 are consistent with the relationship $f \propto \sqrt{T}$ where $T$ is the tension in the nylon string. The nylon string used has a density of $1150 \text{ kg m}^{-3}$ and a uniform diameter of $5.0 \times 10^{-3} \text{ m}$. Determine the length l of the string used. $l =$ The student uses the relationship in question 02.1 to predict frequencies for tensions that are much larger than those used in the original experiment. Explain how the actual frequencies produced would be different from those that the student predicts.

Photo AI

Figure 1 shows an arrangement used by a student to investigate vibrations in a stretched nylon string of fixed length l - AQA - A-Level Physics - Question 2 - 2017 - Paper 1

Question 2

Figure-1-shows-an-arrangement-used-by-a-student-to-investigate-vibrations-in-a-stretched-nylon-string-of-fixed-length-l-AQA-A-Level Physics-Question 2-2017-Paper 1.png

Figure 1 shows an arrangement used by a student to investigate vibrations in a stretched nylon string of fixed length l. He measures how the frequency f of first-har... show full transcript

Worked Solution & Example Answer:Figure 1 shows an arrangement used by a student to investigate vibrations in a stretched nylon string of fixed length l - AQA - A-Level Physics - Question 2 - 2017 - Paper 1

Step 1

Show that the data in Table 1 are consistent with the relationship $f \propto \sqrt{T}$

96%

114 rated

Answer

To verify the given relationship, we will calculate the tension $T$ for each mass using the formula:

$T = mg$

where $m$ is the mass and $g$ is the acceleration due to gravity (approximately $9.81 \text{ m/s}^2$ ). For each mass from Table 1:

For $m = 0.50$ kg:
$T = 0.50 \times 9.81 = 4.905 \text{ N}$
For $m = 0.80$ kg:
$T = 0.80 \times 9.81 = 7.848 \text{ N}$
For $m = 1.20$ kg:
$T = 1.20 \times 9.81 = 11.772 \text{ N}$

Next, we will calculate the square root of the tension:

For $T = 4.905$ :
$\sqrt{T} = \sqrt{4.905} \approx 2.22$
For $T = 7.848$ :
$\sqrt{T} = \sqrt{7.848} \approx 2.80$
For $T = 11.772$ :
$\sqrt{T} = \sqrt{11.772} \approx 3.43$

Now, we will determine the ratio of frequencies to the square roots of the tensions:

For $m = 0.50$ kg:
$\frac{f}{\sqrt{T}} = \frac{110}{2.22} \approx 49.55$
For $m = 0.80$ kg:
$\frac{f}{\sqrt{T}} = \frac{140}{2.80} \approx 50.00$
For $m = 1.20$ kg:
$\frac{f}{\sqrt{T}} = \frac{170}{3.43} \approx 49.54$

Since the ratios are approximately the same, the data is consistent with the relationship $f \propto \sqrt{T}$ .

Step 2

Determine the length l of the string used.

99%

104 rated

Answer

To find the length $l$ of the string used, we start with the formula relating the tension $T$ , the mass per unit length $\mu$ , and the frequency $f$ :

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

Here, we know that:

The density of the string is $1150 \text{ kg/m}^3$ .
The diameter of the string is $5.0 \times 10^{-3} \text{ m}$ .

First, we calculate the cross-sectional area $A$ of the string:

$A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{5.0 \times 10^{-3}}{2}\right)^2 \approx 1.9635 \times 10^{-5} \text{ m}^2$

Next, we find the mass per unit length $\mu$ :

$\mu = \text{density} \times A = 1150 \times 1.9635 \times 10^{-5} \approx 0.0226 \text{ kg/m}$

Using the tension $T$ computed earlier (for example, from $m = 1.20$ kg, $T = 11.772 \text{ N}$ ) and the frequency corresponding to this mass, $f = 170 \text{ Hz}$ , we can substitute in to find $l$ :

$170 = \frac{1}{2l} \sqrt{\frac{11.772}{0.0226}}$

Rearranging gives:

$l = \frac{1}{2} \frac{\sqrt{0.0226}}{\sqrt{11.772}}$

Calculating this yields:

$l \approx 0.067 \text{ m}$

Step 3

Explain how the actual frequencies produced would be different from those that the student predicts.

96%

101 rated

Answer

When predicting frequencies for tensions that are significantly larger than those used in the original experiment, it is important to consider the effect of increased tension on the behavior of the nylon string. As tension increases, it can have a couple of notable effects:

Non-Linearity: The predictions based on the linear relationship $f \propto \sqrt{T}$ assume that the string behaves linearly, which may not hold true at higher tensions.
Material Properties: The nylon string could experience changes in elasticity or tension distribution at much higher tensions, which can affect frequency.
Damping Effects: At higher frequencies and tensions, energy losses due to damping can be increased, further reducing the actual observed frequency compared to theoretical predictions.

Consequently, the actual frequencies observed at higher tensions may be lower than those predicted by the model, making the predictions less valid.

Figure 1 shows an arrangement used by a student to investigate vibrations in a stretched nylon string of fixed length l - AQA - A-Level Physics - Question 2 - 2017 - Paper 1

Worked Solution & Example Answer:Figure 1 shows an arrangement used by a student to investigate vibrations in a stretched nylon string of fixed length l - AQA - A-Level Physics - Question 2 - 2017 - Paper 1

Show that the data in Table 1 are consistent with the relationship $f \propto \sqrt{T}$

Determine the length l of the string used.

Explain how the actual frequencies produced would be different from those that the student predicts.

Join the A-Level students using SimpleStudy...