A 1.5 m long pipe has one end open. What is the length of the first harmonic, i.

Question

A 1.5 m long pipe has one end open. What is the length of the first harmonic, i.e., the longest wavelength among the possible standing-wave which could be generated in this pipe? The relation between the length of pipe L and the wavelengths is: L = 1/4 lambda_ 1/4 times V/f_1 and L = 3/4 lambda_2 = 5/4 lambda_2 = = (2 - 1)/4 lambda_ = (2 - 1)/4 times v/f_ Compare the two equations above, we have 1/4 times V/f_ = (2m - 1)/4 times v/f_ Compare the two sides of this equation, we have 1/f_ = (2m - 1)/f_ Therefore f_m = (2m - 1) middot f_1, where f_m is the frequency of the m^th standing wave. If among the possible standing-waves f_m = 225 Hz and f_m + 1= 375 Hz. What is the fundamental frequency f_1, and the sound speed v? f_m = (2m - 1)middot f_1 = 225 f_m + 1 = [2(m + 1) - 1] middot f_1 = 375

Explanation / Answer

fm = (2m-1)*f1 = 225

fm+1 = (2(m+1) -1)f1 =375

         = (2m+1)f1 = 375      

(2m+1)f1 - (2m-1)f1 =150

2f1 = 150 ,f1 = 75 , fundamental frequency

L = V/4f1 = 1.5m

V = 1.5*300 = 450 m/s

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