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A simple apparatus for demonstrating resonance in an air column is depicted in Fig. 7.59.A vertical pipe open at both ends is partially submerge in water , and a tuning fork vibration at an unknown frequency is placed near the top of the pipe . The length L of the air column can be adjusted by moving the pipe vertically . The sound waves generated by the fork are reinforced of the pipe . For a certain pipe , the smallest value of L for which a peak occurs in the intensity is 9.00 cm. a. What is the frequency of the tuning fork ? b. What are the values of L for the next two resonance conditions? |
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Answer» Solution :Consider how this problem differs from the PRECEDING problem. In the culvert , the length was fixed and the air column was presented with a mixture of many frequencies . The pipe in this exampleis presented with one single frequency from the tuning fork , and the lenght of the pipe is varied until RESONANCE is achieved. This EXAMPLE is a simple substitution problem . Although the pipe is open at its lower end to allow the water to enter , the water's surface acts like a barrier . Therefore , this setup can be modelled as an air column closed at one end . The fundamental frequency for `L =0.090 m`: `f_(1) = (V)/( 4L) = ( 343 m//s )/(( 4( 0.090 m)) = 953 Hz` Because the tuning fork causes the air column to resonate at this frequency , this frequency must also be that of the tuning fork. b. Use equation `v = lambda f` to FIND the wavelength of the sound wave from the tuning fork. `lambda = (v)/(f) = ( 343 m//s)/( 953 Hz) = 0.360 m` Notice from Fig. 7.59 (b) that the length of the air column for the second resonance is ` 3 lambda//4`. `L = 3 lambda//4 = 0.270 m` Notice from Fig. 7.59(b) that the length of the air column for the third resonance is ` 5 lambda//4` `L = 5 lambda//4 = 0.450 m` |
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