First slide

Standing waves

Question

Two wires are fixed in a sonometer. Their tensions are in the ratio 8 : 1. The lengths are in the ratio 36 : 35. The diameters are in the ratio 4 : 1. Densities of the materials are in the ratio 1 : 2. If the higher frequency in the setting is 360 Hz, the beat frequency when the two wires are sounded together is

Moderate

A

10
B

6
C

8
D

5

Solution

Let I, d, f, p and p be the length, diameter, tension, density (mass per unit length] and number of loops of sonometer wire respectively. The frequency n of a stretched string is given by
$n = \frac{p}{Ld} \sqrt{(\frac{T}{ρ})} or n \propto \frac{1}{Ld} \sqrt{(\frac{T}{ρ})} \frac{n_{1}}{n_{2}} = \sqrt{(\frac{T_{1} ρ_{2}}{T_{2} ρ_{1}})} \times \frac{L_{2}}{L_{1}} \times \frac{d_{2}}{d_{1}} = \sqrt{(\frac{8}{1}) (\frac{2}{1})} \times \frac{35}{36} \times \frac{1}{4} = 0 \cdot 972 Now \frac{n_{2}}{360} = 0 \cdot 972 or n_{2} = 350 Beat frequency = 360 - 350 = 10$

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