A certain organ pipe, open at both ends, produces a fundamental frequency of 262

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A certain organ pipe, open at both ends, produces a fundamental frequency of 262Hz in air. Part A If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.8g/mol and the molar mass of helium to be 4.00g/mol . Express your answer in hertz. fHe =   Hz   SubmitHintsMy AnswersGive UpReview Part
A certain organ pipe, open at both ends, produces a fundamental frequency of 262Hz in air. Part A If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.8g/mol and the molar mass of helium to be 4.00g/mol . Express your answer in hertz. fHe =   Hz   SubmitHintsMy AnswersGive UpReview Part
A certain organ pipe, open at both ends, produces a fundamental frequency of 262Hz in air. A certain organ pipe, open at both ends, produces a fundamental frequency of 262Hz in air. A certain organ pipe, open at both ends, produces a fundamental frequency of 262Hz in air. Part A If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.8g/mol and the molar mass of helium to be 4.00g/mol . Express your answer in hertz. fHe =   Hz   SubmitHintsMy AnswersGive UpReview Part
Part A If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.8g/mol and the molar mass of helium to be 4.00g/mol . Express your answer in hertz. fHe =   Hz   SubmitHintsMy AnswersGive UpReview Part
Part A If the pipe is filled with helium at the same temperature, what fundamental frequency fHe will it produce? Take the molar mass of air to be 28.8g/mol and the molar mass of helium to be 4.00g/mol . Express your answer in hertz. fHe =   Hz   SubmitHintsMy AnswersGive UpReview Part
fHe =   Hz   fHe =   Hz   SubmitHintsMy AnswersGive UpReview Part

Explanation / Answer

Let p = density.

p(air) = 1.4, p(He) = 5/3

Sound velocity c = sqrt(p*pressure/p) (see the ref.)

Frequency f is proportional to c, and pressure is the same in both cases, thus

f(He) = f(air) * sqrt[(p(He)/p(air)) * (p(air)/p(He))]

         = 262 * sqrt[(5/3)/1.4 * 28.8/4]

          = 767 Hz

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