A student is investigating forced vertical oscillations in springs. Two springs, A and B, are suspended from a horizontal metal rod that is attached to a vibration generator. The stiffness of A is k, and the stiffness of B is 3k. Two equal masses are suspended from the springs as shown in Figure 10. The vibration generator is connected to a signal generator. The signal generator is used to vary the frequency of the vibration of the metal rod. When the signal generator is set at 2.0 Hz, the mass attached to spring A oscillates with a maximum amplitude of $2.5 \times 10^{-2}$ m and has a maximum kinetic energy of 54 mJ. 1. Deduce the spring constant k for spring A and the mass m suspended from it. 2. Calculate the frequency at which the mass attached to spring B oscillates with maximum amplitude. frequency = ___ Hz 3. Figure 11 shows how the amplitude of the oscillations of the mass varies with frequency for spring A. The investigation is repeated with the mass attached to spring B immersed in a beaker of oil. A graph of the variation of the amplitude with frequency for spring B is different from the graph in Figure 11. Explain two differences in the graph for spring B.

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A student is investigating forced vertical oscillations in springs - AQA - A-Level Physics - Question 7 - 2017 - Paper 1

Question 7

A student is investigating forced vertical oscillations in springs. Two springs, A and B, are suspended from a horizontal metal rod that is attached to a vibration g... show full transcript

Worked Solution & Example Answer:A student is investigating forced vertical oscillations in springs - AQA - A-Level Physics - Question 7 - 2017 - Paper 1

Step 1

1. Deduce the spring constant k for spring A and the mass m suspended from it.

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Answer

To find the spring constant k, we can use the formula for maximum kinetic energy:

$KE = \frac{1}{2} k A^2$

Given that $KE = 54 \text{ mJ} = 54 \times 10^{-3} \text{ J}$ and $A = 2.5 \times 10^{-2} \text{ m}$ , we can rearrange to find k:

$54 \times 10^{-3} = \frac{1}{2} k (2.5 \times 10^{-2})^2$

Solving for k:

$k = \frac{2 \times 54 \times 10^{-3}}{(2.5 \times 10^{-2})^2} = \frac{0.108}{6.25 \times 10^{-4}} = 172.8 \text{ N m}^{-1}$

For the mass m, using the formula for the angular frequency:

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Rearranging to find m:

$m = \frac{k}{(2\pi f)^2}$

Substituting k and f = 2.0 Hz:

$m = \frac{172.8}{(2\pi \cdot 2)^2} = \frac{172.8}{16\pi^2} \approx 0.33 \text{ kg}$

Step 2

2. Calculate the frequency at which the mass attached to spring B oscillates with maximum amplitude.

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Answer

Since the stiffness of spring B is 3k, we find:

$k_B = 3 \times 172.8 \approx 518.4 \text{ N m}^{-1}$

Using the same mass m (0.33 kg), we calculate the frequency:

$f_B = \frac{1}{2\pi} \sqrt{\frac{k_B}{m}} = \frac{1}{2\pi} \sqrt{\frac{518.4}{0.33}} \approx 3.0 \text{ Hz}$

Step 3

3. Explain two differences in the graph for spring B.

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Answer

The peak amplitude for spring B will be at a higher frequency compared to spring A due to the increased stiffness of spring B, which allows for resonant oscillations at a higher frequency.
The width of the resonance peak for spring B will be narrower than that of spring A, indicating less damping effect on oscillations, leading to a sharper peak in the graph.

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A student is investigating forced vertical oscillations in springs - AQA - A-Level Physics - Question 7 - 2017 - Paper 1

Worked Solution & Example Answer:A student is investigating forced vertical oscillations in springs - AQA - A-Level Physics - Question 7 - 2017 - Paper 1

1. Deduce the spring constant k for spring A and the mass m suspended from it.

2. Calculate the frequency at which the mass attached to spring B oscillates with maximum amplitude.

3. Explain two differences in the graph for spring B.

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