Two open organ pipes of lengths 50 cm and 50.6 cm produces 4 beats/s. The speed of sound in air is about-

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Solution:

For an open organ pipe, the fundamental frequency is given by $f = \frac{v}{2 L}$ .

Given $L_{1} = 50 c m = 0.50 m$ and $L_{2} = 50.6 c m = 0.506 m$ .

Beat frequency $f_{b} = ∣ f_{1} - f_{2} ∣ = 4 Hz$ .

$f_{b} = ∣ \frac{v}{2 L _{1}} - \frac{v}{2 L _{2}} ∣ = v ∣ \frac{1}{2 L _{1}} - \frac{1}{2 L _{2}} ∣ = \frac{v}{2} ∣ \frac{1}{L _{1}} - \frac{1}{L _{2}} ∣$

$4 = \frac{v}{2} ∣ \frac{1}{0.50} - \frac{1}{0.506} ∣$

$8 = v ∣ \frac{0.506 - 0.50}{0.50 \times 0.506} ∣$

$8 = v \frac{0.006}{0.253}$

$v = \frac{8 \times 0.253}{0.006} = \frac{2.024}{0.006} \approx 337.33 m / s$ .

The closest option is $340 m / s$ .

The final answer is $340 m / s$ .

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