Question
4. Agrade 11 physics teacher wants to demonstrate interference of sound waves for her students so she sets up two speakers 0.75 m apart that are both playing a tone of 340 Hz. The speakers are operating in phase. Determine the wavelength of the sound wave if the speed of sound in the class is 345m/s (1) b. A student standing 6.2 m away from one speaker and 4.7 m away from the other speaker hears only a very quiet sound. Which nodal line are they on? (2) c.What angle does this nodal point make with the central bisector? (1)
Solution
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Blaise
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Answer
**a. Determining the wavelength:**The relationship between wavelength (λ), frequency (f), and wave speed (v) is given by:v = fλWe are given:* v = 345 m/s* f = 340 HzWe can solve for λ:λ = v / fλ = 345 m/s / 340 Hzλ = 1.015 m (approximately)**b. Identifying the nodal line:**The path difference between the waves reaching the student is |6.2 m - 4.7 m| = 1.5 m.Destructive interference (resulting in a quiet sound or a node) occurs when the path difference is an odd multiple of half the wavelength (λ/2). Let's see which nodal line this corresponds to:Path difference = (2n + 1) * (λ/2) , where n = 0, 1, 2, 3... represents the nodal line number.1.5 m = (2n + 1) * (1.015 m / 2)3 m = (2n + 1) * 1.015 m2n + 1 = 3 m / 1.015 m 2n + 1 ≈ 2.9562n ≈ 1.956n ≈ 0.978Since n must be an integer, we round n to the nearest integer, which is 1. Therefore, the student is standing on the **first nodal line (n=1)**.**c. Calculating the angle:**The angle (θ) that a nodal line makes with the central bisector can be approximated using the following formula for small angles:sin θ ≈ tan θ = (Path difference) / (Distance between speakers)In our case:tan θ = 1.5 m / 0.75 m = 2θ = arctan(2) θ ≈ 63.4 degreesTherefore, the nodal point makes an angle of approximately **63.4 degrees** with the central bisector.