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Alla [95]
3 years ago
12

A bugle can be thought of as an open-end pipe. If a bugle were straightened out, it would be 2.65 m long. a. If the speed of sou

nd is 343 m/s, find the lowest frequency (i.e., the largest wavelength) that can resonate in a bugle. (64.72 Hz) b. Find the next two higher resonant frequencies in the bugle. What harmonics are they
Physics
1 answer:
vitfil [10]3 years ago
5 0

Answer

given,

Length of pipe, L = 2.65 m

speed of sound, v = 343 m/s

Pipe is open end Pipe

a) Lowest frequency

  condition, for open end pipe

     \lambda_1 = 2 L

     \lambda_1 = 2\times 2.65  

     \lambda_1= 5.30 m

   For lowest frequency

    f_1 = n\dfrac{v}{\lambda_1}

     n = 1

    f_1 = \dfrac{343}{5.30}

           f₁ = 64.72 Hz

b) For second frequency in bugle

        n = 2

   f_2 =2\times \dfrac{343}{5.30}

          f₂ = 129.43 Hz

   for n = 3

   f_3 =3\times \dfrac{343}{5.30}

          f₃ = 194.16 Hz

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Pluging known information to get :

a = (vf^2-vi^2)/2(xf-xi)

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P=1/2*m*vf^2-1/2*m*vi^2/Δt

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