Question #64440

i needed to ask 2 questions real fast please cuz i dont get how to solve:
1-An open tube 34 cm long resonates in its second overtone with a tuning fork. The air temperature in the room is 15 degrees. Calculate the wavelength of the sound waves and the frequency of the tuning fork.
2-An open organ pipe has a fundamental frequency of 212 Hz at room temperature 20 degrees. What is the length of the pipe?
Please explain what i have to do for both. I am really confused. Thanks

Expert's answer

Answer on question #64440, Physics / Other

Question 1-An open tube 34 cm long resonates in its second overtone with a tuning fork. The air temperature in the room is 15 degrees. Calculate the wavelength of the sound waves and the frequency of the tuning fork. 2-An open organ pipe has a fundamental frequency of 212 Hz at room temperature 20 degrees. What is the length of the pipe?

Solution 1. The fundamental frequency is connected to length of tube as

f0=c4Lf_{0}=\frac{c}{4L}

where cc is speed of sound. Here we suppose that tube is one-end-opened. The speed of sound is

c=(331.3+0.606ϑ)c=(331.3+0.606\cdot\vartheta)

where ϑ\vartheta is the temperature in degrees Celsius. Hence

c=331.1+0.60615=340.19m/sc=331.1+0.606\cdot 15=340.19\,m/s

Hence, fundamental frequqency is

f=340.190.344250Hzf=\frac{340.19}{0.34\cdot 4}\approx 250\,Hz

Then, second overtone is

2504=1000Hz250\cdot 4=1000\,Hz

The correspondent wave lenght is

λ=cν=3401000=0.34m\lambda=\frac{c}{\nu}=\frac{340}{1000}=0.34\,m

2. Again lets first find speed of sound

c=331.1+0.60620=343.2m/sc=331.1+0.606\cdot 20=343.2\,m/s

Then length is

L=c4f0=343.242120.40mL=\frac{c}{4f_{0}}=\frac{343.2}{4\cdot 212}\approx 0.40\,m

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