Question

1 4 (D PUNTOS) PROBLEMA ( S) Una persona encuentra un pozo de una mina abandonada y desea medir su profundidad. Utilizando un oscilador de audio de frecuencia variable observa que se producen en el pozo resonancias sucesivas a frecuencias de 63.58 Hz y 89.25 Cual es la profundidad del pozo? +16.earn) a,ー畾 843 o568 지-ễn.ag

A person finds a pit en an abandoned mine and wants to measure its depth. Using an audio oscilator of variable frecuency he observes that succesive resonances are produced at frecuencies of 63.58HZ and 89.25 HZ. What is the depth of the pit? speed of sound is 89.25HZ

Answer 1

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It is equivalent to closed organ pipe i.e pipe open at one end while closed at other end. The frequency of normal modes of vibration V_m = (2m - 1) V/4 l Because only odd harmonics are formed Let the two consecutive frequencies are for m = n and m = n + 1 V_n = (2n - 1) V/4 l = 63.58 V_n + 1 = (2n + 1) V/4l = 89.25 Therefore 2n + 1/2n - 1 = 89.25/63.58 127.16 n + 63.58 = 178.5n - 89.25 (178.5 - 127.16)n = 89.25 + 63.58 51.34 n = 152.83 n = 2.98 almostequalto 3 Therefore (6 - 1) 343/4 l = 63.58 l = 5 times 343/4 times 63.58 = 6.74 m