In the figure below, S is a small loudspeaker driven by an audio oscillator with a frequency that is varied from 1000 Hz to 2000 Hz, and D is a cylindrical pipe with two open ends and a length of 55.0 cm. The speed of sound in the air-filled pipe is 344 m/s.
(In the figure s is a little speaker next to the pipe D)
(a) At how many frequencies does the sound from the loudspeaker set up resonance in the pipe?
? frequencies
(b) What is the lowest frequency at which resonance occurs?
? Hz
(c) What is the second lowest frequency at which resonance occurs?
? Hz
Isn't the pipe length equal to 1/2 wavelenght?
n*freq(2*length)=speedsound
The lowest frequency will be n=1
the second lowest will be n=2
Ignore this post. My mind is tired.
To determine the number of frequencies at which resonance occurs in the pipe, we need to consider the conditions for resonance in a cylindrical pipe with two open ends.
Resonance in a pipe occurs when the length of the pipe is an integer multiple of half the wavelength of the sound wave. In other words, the distance between the two open ends of the pipe should be equal to a multiple of half the wavelength of the sound wave.
The wavelength of a sound wave can be determined using the formula:
λ = v/f
Where λ is the wavelength, v is the speed of sound, and f is the frequency of the sound wave.
Given that the speed of sound in the air-filled pipe is 344 m/s and the length of the pipe is 55.0 cm (or 0.55 m), we can calculate the minimum and maximum wavelengths that can resonate in the pipe.
(a) At how many frequencies does the sound from the loudspeaker set up resonance in the pipe?
To find the number of frequencies at which resonance occurs, we need to determine the range of possible wavelengths that can satisfy the conditions for resonance in the pipe.
The minimum wavelength occurs when the length of the pipe is half the wavelength:
0.55 m = λ/2
Solving for λ, we get:
λ = 0.55 m * 2 = 1.1 m
The maximum wavelength occurs when the length of the pipe is equal to the wavelength:
0.55 m = λ
Therefore, the range of possible wavelengths is from 1.1 m to 0.55 m.
Now, we can calculate the frequency range for resonance in the pipe by using the formula:
f = v/λ
For the minimum frequency, we use the maximum wavelength:
f_min = 344 m/s / 1.1 m ≈ 312.73 Hz
For the maximum frequency, we use the minimum wavelength:
f_max = 344 m/s / 0.55 m ≈ 625.45 Hz
So, the sound from the loudspeaker can set up resonance in the pipe at frequencies ranging from approximately 312.73 Hz to 625.45 Hz.
(b) What is the lowest frequency at which resonance occurs?
The lowest frequency at which resonance occurs is the minimum frequency we calculated:
f_min ≈ 312.73 Hz
(c) What is the second lowest frequency at which resonance occurs?
To find the second lowest frequency, we need to calculate the frequency that corresponds to the second lowest wavelength.
To find the wavelength, we can use the formula:
λ = 2L/n
Where L is the length of the pipe (0.55 m) and n is the mode of resonance. The mode of resonance refers to the number of half-wavelengths that fit within the pipe length.
For the second lowest resonance mode, n = 2:
λ_2 = 2 * 0.55 m / 2 = 0.55 m
Using the formula f = v/λ, we can calculate the frequency:
f_2 = 344 m/s / 0.55 m ≈ 625.45 Hz
Therefore, the second lowest frequency at which resonance occurs is approximately 625.45 Hz.