A student uses an audio oscillator of adjustable frequency to measure the depth

Question

A student uses an audio oscillator of adjustable frequency to measure the depth of a water well. The student reports hearing two successive resonances at 51.55 Hz and 61.70 Hz. (Assume that the speed of sound in air is 343 m/s.) How deep is the well? Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out ail intermediate results to at leas four-digit accuracy to minimize roundoff error. m How many antinodes are in the standing wave at 51.55 Hz? Your response differs from the correct answer by more than 10%. Double check your calculations. antinodes

Explanation / Answer

A) difference between successive frequencies is v/2l = 61.7-51.55 =10.15 Hz

v/2l = 10.15

v = 343 m/s

depth of the well is l = v/(2*10.15) = 343/(2*10.15) = 16.9 m

B) let n be the number of antinodes then

(n+1)*f= 51.55

f be the fundamental frequency f = v/(4l) = 0.5*10.15 = 5.075 Hz

(n+1)*5.075 = 51.55

n+1 = 10.15

n = 10.15-1 = 9.15

So then umber of anti nodes is 9

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.