Off-tune tuning fork

Two tuning forks with slightly different lengths l 1 = 10 cm l_1 = 10 \, \text{cm} and l 2 = 10.05 cm l_2 = 10.05 \,\text{cm} are vibrated simultaneously. Because their vibration frequencies differ, an acoustic beating is created, so that the sound gets loud and quiet twice a second. What is the natural frequency f 1 f_1 of the longer tuning fork in units of hertz?


The answer is 402.

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1 solution

Markus Michelmann
Dec 18, 2017

The natural vibrations of the tuning forks are standing waves with sound velocity c = λ f c = \lambda f . For the fundamental vibration, the wavelength is λ = 4 l \lambda = 4 l , since the top corresponds to anti-node at and the bottom corresponds to a node, so that the oscillation frequency is f = 1 4 c / l f = \frac{1}{4} c / l .

The superimposition of two sine waves with the frequencies f 1 f_1 and f 2 f_2 results sin ( 2 π f 1 t ) + sin ( 2 π f 2 t ) = sin ( π ( f + + f ) t ) + sin ( π ( f + f ) t ) = [ sin ( π f + t ) cos ( π f t ) + cos ( π f + t ) sin ( π f t ) ] + [ sin ( π f + t ) cos ( π f t ) cos ( π f + t ) sin ( π f t ) ] = 2 sin ( π f + t ) cos ( π f t ) \begin{aligned} \sin(2 \pi f_1 t) + \sin (2 \pi f_2 t) &= \sin(\pi (f_+ + f_-) t) + \sin(\pi (f_+ - f_-) t)\\ &= [\sin(\pi f_+ t)\cos(\pi f_- t) + \cos(\pi f_+ t)\sin(\pi f_- t)] \\ &\quad + \,[\sin(\pi f_+ t)\cos(\pi f_- t) - \cos(\pi f_+ t)\sin(\pi f_- t)] \\ &= 2 \sin(\pi f_+ t)\cos(\pi f_- t) \end{aligned} with f + = f 1 + f 2 f_+ = f_1 + f_2 and f = f 1 f 2 f_- = f_1 - f_2 . The result is a sine wave with the mean frequency f + / 2 f_+/ 2 whose amplitude is modulated by a cosine oscillation with the frequency f / 2 f_- / 2 . This phenomenon is also called (acoustic) beating.

The amplitude is maximized for the times t = n / f = n Δ t t = n / f_- = n \Delta t , n = 0 , 1 , 2 , n = 0,1,2, \dots , so that Δ t = 0.5 s \Delta t = 0.5 \, \text{s } yields a difference frequency f = 2 Hz f_- = 2 \, \text{Hz} . It follows f = c 4 ( 1 l 1 1 l 2 ) c = 4 f 1 l 1 1 l 2 f 1 = c 4 l 1 = f 1 l 1 l 2 = 402 Hz \begin{aligned} f_- &= \frac{c}{4} \left(\frac{1}{l_1} - \frac{1}{l_2} \right) \\ \Rightarrow \quad c &= \frac{4 f_-}{\frac{1}{l_1} - \frac{1}{l_2}} \\ \Rightarrow \quad f_1 &= \frac{c}{4 l_1} = \frac{f_-}{1 - \frac{l_1}{l_2}} = 402\,\text{Hz} \end{aligned}

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