Question #219381

The speed of sound is 340 m/s where a tuning fork produces the second resonance position above a closed air column that is 49.8 cm in length. The frequency of the tuning fork is ___ Hz. (Give your answer with the correct number of sig digs and do not include units in your answer)



1
Expert's answer
2021-07-21T09:06:26-0400

The second resonance position corresponds to the third harmonic of the closed air column. Therefore, we can find the wavelength of the wave:


λ=43L=430.498 m=0.664 m.\lambda=\dfrac{4}{3}L=\dfrac{4}{3}\cdot0.498\ m=0.664\ m.

Finally, we can find the frequency of the tuning fork:


v=fλ,v=f\lambda,f=vλ=340 ms0.664 m=510 Hz.f=\dfrac{v}{\lambda}=\dfrac{340\ \dfrac{m}{s}}{0.664\ m}=510\ Hz.

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