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 length

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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 length of the flute remains constant,the wavelength $\lambda$ of the fundamental note remains constant. Given $v = n\lambda,$ 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 $T_2 = 273 + 31 = 304\,K.$Then,$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{304}{300}} = \left(1 + \frac{4}{300}ight)^{1/2}.$Using the binomial approximation $(1+x)^n \approx 1 + nx$ for small $x,$ we get $\frac{n_2}{n_1} \approx 1 + \frac{1}{2} 	imes \frac{4}{300} = 1 + \frac{2}{300}.$Thus,$n_2 = n_1 \left(1 + \frac{2}{300}ight) = 300 + 2 = 302\,Hz.$The number of beats per second is $|n_2 - n_1| = |302 - 300| = 2\,Hz.$

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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