Question

Refer to figure 1 and draw the resonance frequency that would be observed for different lengths.

Figure 1: Modes of vibration for a tube closed at one end, TUBE CLOSED AT ONE END (a) Displacement of air The same tube will also First harmonic fundamental resonate atcertain higher frequencies, called harmonics. The harmonic frequencies of a tube with one open and one closed end are given by 4 140 Third harmonic 4 3 340 =n- Overtones Fifth harmonic -5U 5-4€

Answer 1

Given

closed pipe , that is open at one end and closed at other end
as the wave is propagating through the closed pipe, always at the open end an antinode will form(maximum displacement)
and at the closed end the node region (displacement will be minimum)

so the frequencies are as follows

   fn = n v/4L, n = 1,3,5,7,9,...... the odd numbers

we know that V = lambda*f ==> f = v/ Lambda

here lambda is wavelength of the wave , given by for the first harmonic

   length L = lambda/4

for second harmonic (first overtone)

   length L = 3*lambda/4
for third harmonic (second overtone)

   length L = 5*lambda/4

for fourth harmonic (third overtone)
   length L = 7*lambda/4

and goes on