A pipe that is closed at one end can be made to resonate by a tuning fork at a length
of 0. 25 m. The next resonant length is 0. 75 m. Speed of sound is 338 m/s.
(a) Calculate the wavelength of the sound emitted by the tuning fork;
(b) Calculate the frequency of the tuning fork.
closed end pipe
lowest res when node at closed end, antinode at open so
1/4 wavelength = 0.25 m
next time node at wall and antinode at open end is when
3/4 wavelength = 0.75 m
well, that works, 3 times .25 is .75 :)
so if L is wavelength
(1/4 ) L = .25 = 1/4
looks like L = 1 meter
it goes one meter in one period T
so 338 T = 1 meter
T = 1/338 seconds
f = 1/T =338 Hertz
To calculate the wavelength of the sound emitted by the tuning fork, we can use the formula:
λ = 2L/n
where:
λ = wavelength
L = length of the pipe
n = harmonic number (i.e., the resonant length divided by the fundamental resonant length)
(a) For the first resonant length of 0.25 m:
λ = 2(0.25 m)/1
λ = 0.50 m
Therefore, the wavelength of the sound emitted by the tuning fork is 0.50 m.
To calculate the frequency of the tuning fork, we can use the formula:
v = fλ
where:
v = speed of sound
f = frequency of the tuning fork
λ = wavelength
Since the speed of sound is given as 338 m/s and the wavelength is 0.50 m (from the previous calculation), we can rearrange the formula to solve for the frequency:
f = v/λ
f = 338 m/s / 0.50 m
f = 676 Hz
Therefore, the frequency of the tuning fork is 676 Hz.
To solve this problem, we need to use the formulas related to the speed of sound, the wavelength, and the frequencies of the resonances in a closed pipe.
Let's break down the problem step by step:
(a) Calculate the wavelength of the sound emitted by the tuning fork:
1. We know that the first resonant length is 0.25 m and the next resonant length is 0.75 m.
2. The fundamental frequency (first harmonic) produced by a closed pipe is equal to four times the length of the pipe (λ = 4L).
Therefore, the wavelength of the sound wave that resonates with a length of 0.25 m is given by the formula:
λ₁ = 4 × 0.25 = 1 m.
3. Similarly, the wavelength associated with the next resonant length of 0.75 m is:
λ₂ = 4 × 0.75 = 3 m.
(b) Calculate the frequency of the tuning fork:
1. We know that the speed of sound is 338 m/s.
2. The speed of sound is equal to the product of the wavelength and the frequency (v = λf).
3. To find the frequency, we can rearrange the formula as follows:
f = v / λ.
4. For the first resonant length, using λ₁ = 1 m:
f₁ = 338 m/s / 1 m = 338 Hz.
5. For the next resonant length, using λ₂ = 3 m:
f₂ = 338 m/s / 3 m ≈ 112.67 Hz.
Therefore, the answers to the questions are:
(a) The wavelength of the sound emitted by the tuning fork is 1 m and 3 m, depending on the resonant length of the pipe.
(b) The frequency of the tuning fork is approximately 338 Hz and 112.67 Hz, corresponding to the resonant lengths.