Question

What are the first 3 resonant frequencies on a 20cm long tube, with one end closed? Use your speed of sound measurement in this calculation.

Speed of sound measurement = 339.8 m/s

Answer 1

Given

the length of the closed tube is L = 0.2 m

speed of sound in air is v = 339.8 m/s

we know that in closed pipe at the open side always an antinode forms so the possible frequencies are  

f1 = v/4L

second resonance frequency is f2 = 3V/4L = 3f1

third resonance frequency is f3 = 5V/4L = 5f1

now substituting the values

f1 = 339.8 /(4*0.2) Hz = 424.75 Hz

f2 = 3f1 = 3*424.75 Hz = 1274.25 Hz

f3 = 5f1 = 5*424.75 Hz = 2123.75 Hz

these are the first three resonance frequencies of the given closed pipe