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)

**a. Determining the wavelength:**<br /><br />The relationship between wavelength (λ), frequency (f), and wave speed (v) is given by:<br /><br />v = fλ<br /><br />We are given:<br />* v = 345 m/s<br />* f = 340 Hz<br /><br />We can solve for λ:<br /><br />λ = v / f<br />λ = 345 m/s / 340 Hz<br />λ = 1.015 m (approximately)<br /><br /><br />**b. Identifying the nodal line:**<br /><br />The path difference between the waves reaching the student is |6.2 m - 4.7 m| = 1.5 m.<br /><br />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:<br /><br />Path difference = (2n + 1) * (λ/2) , where n = 0, 1, 2, 3... represents the nodal line number.<br /><br />1.5 m = (2n + 1) * (1.015 m / 2)<br />3 m = (2n + 1) * 1.015 m<br />2n + 1 = 3 m / 1.015 m <br />2n + 1 ≈ 2.956<br />2n ≈ 1.956<br />n ≈ 0.978<br /><br />Since 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)**.<br /><br />**c. Calculating the angle:**<br /><br />The angle (θ) that a nodal line makes with the central bisector can be approximated using the following formula for small angles:<br /><br />sin θ ≈ tan θ = (Path difference) / (Distance between speakers)<br /><br />In our case:<br /><br />tan θ = 1.5 m / 0.75 m = 2<br /><br />θ = arctan(2) <br />θ ≈ 63.4 degrees<br /><br />Therefore, the nodal point makes an angle of approximately **63.4 degrees** with the central bisector.<br />