First slide

Standing waves

Question

Two uniform wires are vibrating simultaneously in their fundamental nodes. The tensions, lengths, diameters and densities of two wires are in the ratio I : 1, 36 : 35, 4 : 1 and 1 : 2 respectively. If the note of the higher pitch has a frequency 360 Hz, the number of beats produced per second is

Moderate

A

5
B

10
C

15
D

20

Solution

we know that
$n = \frac{1}{LD} \sqrt{(\frac{T}{ρ})} \frac{n_{1}}{n_{2}} = \frac{L_{2} d_{2}}{L_{1} d_{1}} \sqrt{(\frac{T_{1}}{ρ_{1}} \times \frac{ρ_{2}}{T_{2}})} = (\frac{L_{2}}{L_{1}}) (\frac{d_{2}}{d_{1}}) \sqrt{(\frac{T_{1}}{T_{2}}) \times (\frac{ρ_{2}}{ρ_{1}})} or \frac{n_{1}}{360} = (\frac{36}{35}) \times (\frac{1}{4}) \sqrt{(\frac{8}{1}) \times (\frac{2}{1})} n_{1} = 350 So, number of beats produced / \sec = n_{2} - n_{1} = 10$

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