Question

An air column in a glass tube is open at one end and closed at the other by a movable piston. The air in the tube is warmed above room temperature, and a 375 Hz tuning fork is held at the open end. Resonance is heard when the piston is at a distance d1 = 24.5 cm from the open end and again when it is at a distance d2 = 68.3 cm from the open end. (a) What speed of sound is implied by this data? (b) How far from the open end will the piston be when the next resonance is heard?

Answer 1

Answer:

a. v = 367.5 m/s

b. L = 1.225m

Explanation:

λn=4*L/n

λ=4*L

λ1=4*0.245m=0.98

λ2=4*0.683m=2.732

We're hearing the first and second harmonics for this tube, so for the first harmonic

a.

f' = n*v / 4*L

375Hz'=1*v/4*0.245m

Solve to v

v = 375Hz*4.0*0.245m

v = 367.5 m/s

b.

Now the next resonance is heard is L' the next harmonic is 5 so:

f' = 5*367.5m/s / 4*L

Solve to L'

L = 5*367.5m/s / 375Hz*4

L = 1.225m