The clarinet is a roughly cylindrical instrument of length L that is open to the

Question

The clarinet is a roughly cylindrical instrument of length L that is open to the outside air at the
bell, but closed by the mouthpiece, reed and the player's mouth at the other end
(a) Find expressions for the first four frequencies of standing waves in the clarinet. Your
expressions should include the length, L and the speed of sound, v.
(b) The lowest note played by a standard clarinet is 146.83 Hz. Use this information and the
speed of sound in air, 343 m/s, to find numerical values for the next three frequencies.

5- The clarinet is a roughly cylindrical instrument of length L that is open to the outside air at the bell, but closed by the mouthpiece, reed and the player's mouth at the other end (a) Find expressions for the first four frequencies of standing waves in the clarinet. Your expressions should include the length, L and the speed of sound, v. (b) The lowest note played by a standard clarinet is 146.83 Hz. Use this information and the speed of sound in air, 343 m/s, to find numerical values for the next three frequencies.

Explanation / Answer

when the clarinet vibrates at the lowest frequency

here in the clarinet at the open end a antinode is formed and

at the closed end node is formed

the distane between the node and antinode is L = lambda/4

lambda = wavelength = 4L

frequency fo = V/lambda = v/4L <-----answer

when the clarinet vibrates at the second lowest frequency

here in the clarinet at the open end a antinode is formed and

node is formed at the middle and closed end

thw length of the clarinet is L = lambda/4 + lamba/2 =
3lambda/4

lambda = wavelength = 4L/3

frequency f1 = V/lambda = 3v/4L   <-----answer

similarly for third frequency f2 = 5v/4L <-----answer

part (b)

given fo = 146.83 Hz      <---------answer

f1 = 3*fo = 3*146.83 = 440.49 Hz     <---------answer

f2 = 5*fo = 734.15 Hz   <---------answer

f3 = 7*fo = 1027.81 Hz    <---------answer

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