Question

In this lab all of your measurements were taken at the shortest length of the PVC pipe above the water that caused the air inside the pipe to resonate. Using the frequency of the first tuning fork from your lab, what would be the minimum length of the PVC pipe required to find the next shortest length of PVC pipe above the water level that will cause the air inside the pipe to resonate? The first tuning fork used has a frequency of 392Hz.

Answer 1

given

frequency is f1 = v/Lambda

f1 = V/(4*l)

because the at the open end the antinode forms and at the water end node will form so (1/4)th of a wavelength wave forms .

the next resonance frequency is at length l = (3/4)lambda  

given f1 = 392 Hz

then the length of the aircoloumn is  

392 = 334 /(4*l)

l = 0.213 m

so the wavelength is lambda = 4*0.213 m = 0.852 m

the next frequency is  

f2 = v/((3/4)lambda

that is length of air coloumn is l2 = (3/4)lambda

L2 = (3/4)(0.852)m = 0.639 m

so the shortest length of PVC pipe above the water level that will cause the air inside the pipe to resonate is = 0.639 m