Jeff discovered an old water well in his backyard. He decided to measure its depth. He happened to have an audio oscillator of adjustable frequency. He sends sound waves down to the well and adjusts the frequency until he finally hears two successive resonances at frequencies of 195 Hz and 221 Hz. What is the depth of the well? Take speed of sound as 440 m/s.
To determine the depth of the well, we can use the principle of resonance. Resonance occurs when the frequency of the sound wave matches the natural frequency of the object or medium, leading to an increase in amplitude. In this case, resonance is observed when the sound waves bounce back and forth between the top and bottom of the well.
The distance between two successive resonances is equal to half the wavelength of the sound wave (λ/2). The wavelength, in turn, is related to the speed of sound (v) and the frequency (f) through the formula:
λ = v / f
Given that the speed of sound is 440 m/s, the frequency of the first resonance is 195 Hz, and the frequency of the second resonance is 221 Hz, we can calculate the wavelengths of these frequencies:
λ1 = 440 m/s / 195 Hz
λ2 = 440 m/s / 221 Hz
The difference in the wavelengths of the two frequencies will give us the distance between the successive resonances:
Δλ = λ2 - λ1
Finally, we can determine the depth of the well by multiplying the distance between the successive resonances by the number of half-wavelengths in the well. Since the sound waves have to travel down and back up, the number of half-wavelengths is 2:
Depth = 2 * (Δλ)
Now let's calculate the depth of the well:
λ1 = 440 m/s / 195 Hz = 2.25 m
λ2 = 440 m/s / 221 Hz = 1.99 m
Δλ = λ2 - λ1 = 1.99 m - 2.25 m = -0.26 m
Keep in mind that this negative difference indicates a phase shift, which means the sound wave completed an extra half-wavelength during the measurement. To get the positive difference, we need to take the absolute value:
Δλ = |-0.26 m| = 0.26 m
Depth = 2 * (0.26 m) = 0.52 m
Therefore, the depth of the well is approximately 0.52 meters.