Two identical flutes produce fundamental note | vedclass

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(B) The velocity of sound $v$ is directly proportional to the square root of the absolute temperature $T,$ i.e.,$v \propto \sqrt{T}.$Since the frequen

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$2$

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(B) The velocity of sound $v$ is directly proportional to the square root of the absolute temperature $T,$ i.e.,$v \propto \sqrt{T}.$Since the frequency $n = \frac{v}{\lambda}$ and the length of the flute (and thus the wavelength $\lambda$) remains constant,we have $n \propto v \propto \sqrt{T}.$Let $n_1 = 300 \ Hz$ at $T_1 = 273 + 27 = 300 \ K.$Let $n_2$ be the frequency at $T_2 = 273 + 31 = 304 \ K.$Using the ratio: $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{304}{300}} = \sqrt{1 + \frac{4}{300}}.$Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$: $\frac{n_2}{300} \approx 1 + \frac{1}{2} 	imes \frac{4}{300} = 1 + \frac{2}{300}.$Thus,$n_2 = 300 + 2 = 302 \ Hz.$The number of beats per second is the difference in frequencies: $|n_2 - n_1| = |302 - 300| = 2 \ Hz.$

Two identical flutes produce fundamental notes of frequency $300 \ Hz$ at $27^{\circ}C.$ If the temperature of air in one flute is increased to $31^{\circ}C,$ the number of beats heard per second will be

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