The physics of wind instruments is based on the concept of standing waves. When

Question

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.

Throughout the problem, take the speed of sound in air to be 343m/s .

Part A

Consider a pipe of length 80.0cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe? Express answer in hertz.

Part B

A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now

A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now

higher than before.

Part C

Express your answer in hertz.

Part D

What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?

What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?

All integer multiples of the fundamental frequency

Part E

The same as the length of the open-open pipe

Part F

Express your answer in hertz.

the same as before. lower than before.

higher than before.

Part C

If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?

Express your answer in hertz.

Part D

What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?

What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?

Only the odd multiples of the fundamental frequency Only the even multiples of the fundamental frequency

All integer multiples of the fundamental frequency

Part E

What length of open-closed pipe would you need to achieve the same fundamental frequency f as the open-open pipe discussed in Part A?

What length of open-closed pipe would you need to achieve the same fundamental frequency as the open-open pipe discussed in Part A?

Half the length of the open-open pipe Twice the length of the open-open pipe One-fourth the length of the open-open pipe Four times the length of the open-open pipe

The same as the length of the open-open pipe

Part F

What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?

Express your answer in hertz.

Explanation / Answer

a) v=?*f,

where v=343 m/s,

L=80.0cm =0.8m;

length of standing wave ?=2*L=1.6m (coz both ends are open!),

Thus fundamental frequency is f = v/?

                                             = 343/(2*0.8)

                                             = 214.375 Hz = 214 Hz;

(b)

Since the hole opens the pipe to the pressure of the surrounding air, the standing wave created in the pipe has an antinode near the hole. In other words, the presence of a hole in the pipe reduces the length of the column of air that can oscillate in the pipe.

Consider the formula used in Part A and use the fact that the length of the vibrating column of air is now shorter

f is higher than before (option C ); because L is shorter, hence ? is shorter, v being the same!

(c)

therefore new fundamental frequency is f ' = v/(L)

                                                          = 428.75 Hz = 429 Hz

(d)

Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.

So Answer is option b) Only the even multiples of the fundamental frequency

(e) The frequencies possible in an open closed pipe of length L are given by

fm = m* v /4L

where m = 1,3,5,...

where v is the speed of sound in the air.

So option is a) Half the length of the open-open pipe

(f)

Recall that possible frequencies of standing waves that can be generated in an open closed pipe include only

odd harmonics. Then the first possible harmonic after the fundamental frequency is the third harmonic

f '' = 3v/(4*0.5*L)

= 643 Hz

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