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.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. 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. the same as before.
b. lower than before.
c. higher than before.
3. 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?
4. 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?
4-1. 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.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. 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.

Answer 1

Answer:

1) f = 214 Hz , 2)  answer is c , 3) f = 428 Hz , 4)   f₂ = 428 Hz ,   f₃ = 643Hz

Explanation:

1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore

                L = λ / 2

               λ  = 2L

               λ  = 2 0.8

               λ  = 1.6 m

wavelength and frequency are related to the speed of sound (v = 343 m / s)

                v =λ  f

                f = v / λ  

                f = 343 / 1.6

                f = 214 Hz

2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced

           λ' = 2 L ’

          as L ’<L₀

          λ' <λ₀

          f = v / λ'

          f' > fo

the correct answer is c

3) in this case the length is L = 0.40 m

          λ = 2 0.4 = 0.8 m

          f = 343 / 0.8

          f = 428 Hz

4) the different harmonics are described by the expression

         λ = 2L / n           n = 1, 2, 3

         λ₂ = L

         f₂ = 343 / 0.8

         f₂ = 428 Hz

         λ₃ = 2 0.8 / 3

         λ₃ = 0.533 m

         f₃ = 343 / 0.533

         f₃ = 643 Hz

4,1) as we have two maximums at the ends, all integer multiples are present

       the answer is C

E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested

in this pipe there is a maximum in the open part and a node in the closed part, so the expression

        L = λ / 4

        L = 1.6 / 4

        L = 0.4 m

the answer is C

F) in this type of pipe the general expression is

           λ = 4L / n         n = 1, 3, 5 (2n + 1)

therefore only odd values ​​can produce standing waves

           λ₃ = 4L / 3

           λ₃ = 4 0.4 / 3

           λ₃ = 0.533

           f₃ = 343 / 0.533

           f₃ = 643 Hz