An air-filled pipe is found to have successive harmonics at 980 Hz , 1260 Hz , and 1540 Hz . It is unknown whether harmonics below 980 Hz and above 1540 Hz exist in the pipe. What is the length of the pipe

  L = 0.7 m

Explanation:

This is a resonance exercise, in this case the air-filled pipe is open at both ends, therefore we have bellies at these points.

          λ / 2 = L                       1st harmonic

          λ = L                            2nd harmonic

          λ = 2L / 3                    3rd harmonic

         λ =  2L / n                    n -th harmonic

the speed of sound is related to wavelength and frequencies

           v =λ f

            f = v /λ

we substitute

            f = v n / 2L

the speed of sound in air is v = 343 m / s

suppose that the frequency of f = 980Hz occurs in harmonic n

            f₁ = v n / 2L

            f₂ = v (n + 1) / 2L

            f₃ = v (n + 2) / 2L

we substitute the values

             2 980/343  =  n / L

              2 1260/343 = (n + 1) / L

               2 1540/343 = (n + 2) / L

we have three equations, let's use the first two

              5.714 = n / L

              7.347 = (n + 1) / L

we solve for L and match the expressions

              n / 5,714 = (n + 1) / 7,347

              7,347 n = 5,714 (n + 1)

              n (7,347 -5,714) = 5,714

              n = 5,714 / 1,633

              n = 3.5

as the number n must be integers n = 4 we substitute in the first equation

              L = n / 5,714

              L = 0.7 m

a=d/t

a="triangle"v/t

explanation:

a = acceleration