PROBLEM Figure a shows a simple apparatus fordemonstrating resonance in a tube. A long tube open atboth ends is partially submerged in a beaker of water, anda vibrating tuning fork of unknown frequency is placed nearthe top of the tube. The length of the air column, L, isadjusted by moving the tube vertically. The sound wavesgenerated by the fork are reinforced when the length of theair column corresponds to one of the resonant frequenciesof the tube. Suppose the smallest value of L for which apeak occurs in the sound intensity is 9.00 cm. (a) Withthis measurement, determine the frequency of the tuning fork. (b) Find the wavelength and the nexttwo air-column lengths giving resonance. Take the speed of sound to be 345 m/s.SOLUTION(A) Find the frequency of the tuning fork.= 1, v = 345 m/s, andSubstitute n =L₁= 9.00 x 10-² m.STRATEGY Once the tube is in the water, the setup is the same as a pipe closed at one end. Forpart (a), substitute values for v and L into the one ended harmonics equation with n= 1 and find thefrequency of the tuning fork. (b) The next resonance maximum occurs when the water level is low enoughto allow a second node, which is another half-wavelength in distance. The third resonance occurs whenthe third node is reached, requiring yet another half-wavelength of distance. The frequency in each case isthe same because it's generated by the tuning fork.Add a half-wavelength of distance toL₁ to get the next resonance position.f₁Add another half-wavelength to L₂ toobtain the third resonance position.=(B) Find the wavelength and the next two water levels giving resonance.Calculate the wavelength, using thefact that, for a tube open at one end,= 4L for the fundamental.λ =Water= 4L₁V345 m/s4L₁ 4(9.00 x 10-² cm)=Firstresonance(a) Apparatus for demonstrating the resonance of soundwaves in a tube closed at one end. The length L of the aircolumn is varied by moving the tube vertically while it ispartially submerged in water. (b) The first three resonancesof the system.4(9.00 x 10-2 m):Secondresonance(thirdharmonic)958 Hz= 0.360 mThirdresonance(fifthharmonic)L₂ = L₁ + 1/2 = 0.090 m + 0.180 m = 0.270 mL3 = L₂ + 1/2 = 0.270 m + 0.180 m = 0.450 m PRACTICE ITUse the worked example above to help you solve this problem. Figure a shows a simple apparatus fordemonstrating resonance in a tube. A long tube open at both ends is partially submerged in a beaker ofwater, and a vibrating tuning fork of unknown frequency is placed near the top of the tube. The length ofthe air column, L, is adjusted by moving the tube vertically. The sound waves generated by the fork arereinforced when the length of the air column corresponds to one of the resonant frequencies of the tube.Suppose the smallest value of L for which a peak occurs in the sound intensity is 8.83 cm. Take the speedof sound to be 341 m/s.(a) With this measurement, determine the frequency of the tuning fork.=Hz(b) Find the wavelength and the next two air-column lengths giving resonance.λ =L₂ =L3=ε ε εEXERCISEAn unknown gas is introduced into the aforementioned apparatus using the same tuning fork, and the firstresonance occurs when the air column is 5.93 cm long. Find the speed of sound in the gas.m/ssoundHINTS: GETTING STARTED I I'M STUCK!

Given,The length of the air column is LThe speed of the sound = 341 m/sThe smallest value of L which a peak occurs in the sound intensity is 8.83 cmFrom the pipe closed at one end and open at other end,a) f = v4Lf = 3414×(8.83×10-2)f = 965.46 Hzb) The wavelength and the next two air column lengths giving resonance,λ= 4L = 4×(8.83×10-2)λ=0.3532 mThe next two air column lengths giving resonance,L2 =3λ4 = 3×(0.3532)4L2=0.265 mL3 = 5λ4=5×(0.3532)4L3=0.4415 mNow the first resonance length is 5.93 cm,The speed of sound in the gas,λ=4L λ=4×(5.93×10-2)λ=0.2372 mWe know,v = fλIn the aforementioned example, f = 958 Hztherefore, v = 958 Hz×0.2372 mv =227.24 m/s