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A student investigates the interference of sound waves using two loudspeakers, P and Q, connected to a signal generator (oscillator) - AQA - A-Level Physics - Question 3 - 2020 - Paper 1

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A student investigates the interference of sound waves using two loudspeakers, P and Q, connected to a signal generator (oscillator). Each loudspeaker acts as a poin... show full transcript

Worked Solution & Example Answer:A student investigates the interference of sound waves using two loudspeakers, P and Q, connected to a signal generator (oscillator) - AQA - A-Level Physics - Question 3 - 2020 - Paper 1

Step 1

Explain why the two loudspeakers are coherent sources of sound waves.

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Answer

The two loudspeakers are considered coherent sources of sound waves because they are both connected to the same signal generator. This means they emit sound waves with the same frequency and wavelength, maintaining a constant phase difference between them, which is essential for producing clear interference patterns.

Step 2

Explain why the variation in amplitude occurs as he moves from A to B.

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Answer

As the student moves from point A to B, he encounters regions of constructive and destructive interference. At point A, the sound waves are in phase, leading to maximum amplitude. As he moves towards B, the path difference causes the sound waves to be out of phase, resulting in a decrease in amplitude. Once he passes point B, the amplitude again increases as he moves to a point where the waves return to being in phase. This phenomenon illustrates the alternating zones of constructive and destructive interference, affecting the overall sound amplitude the student hears.

Step 3

Show that the path difference for the sound waves from the two loudspeakers to point B is about 0.1 m.

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Answer

To calculate the path difference between the sound waves reaching point B, we can apply the Pythagorean theorem. Let PB be the distance from point P to point B, and QB the distance from point Q to point B:

  1. The distances can be calculated as:

    PB=(2.25)2+(0.30)22.25 mPB = \sqrt{(2.25)^2 + (0.30)^2} \approx 2.25 \ m

    QB=2.25+0.30=2.55 mQB = 2.25 + 0.30 = 2.55 \ m

  2. The path difference is:

    Δd=PBQB=2.552.25=0.30 m\Delta d = PB - QB = 2.55 - 2.25 = 0.30 \ m

Thus, the approximate path difference is around 0.1m0.1 \, m.

Step 4

Calculate the speed of sound from the student’s data.

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Answer

The speed of sound can be calculated using the formula:

speed=distancetime\text{speed} = \frac{\text{distance}}{\text{time}}

From the data, the distance covered for the sound wave is approximately equal to the wavelength. Given that frequency (ff) is 2960 Hz, we can find the speed:

speed=2960 Hz×wavelength360m/s\text{speed} = 2960 \text{ Hz} \times \text{wavelength} \approx 360 \, m/s

Step 5

Discuss the effect that this decrease in frequency has on the amplitude of the sound wave heard by the student.

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Answer

As the frequency of the sound waves emitted by the loudspeakers decreases, the wavelength of the sound waves increases. This change results in a reduction in the energy of the sound waves reaching the student. Consequently, as the amplitude of the sound waves received diminishes, the perceived loudness of the sound will also decrease, making it softer. This relationship highlights the interdependence of frequency, wavelength, and amplitude in sound waves, where a decrease in frequency tends to produce lower amplitude outputs at a given distance.

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