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deff fn [24]
2 years ago
12

Two cellists, one seated directly behind the other in an orchestra, play the same note for the conductor who is directly in fron

t of them. Because of the separation between the cellists, destructive interference occurs at the conductor. This separation is the smallest that produces destructive interference. Would this separation increase, decrease, or remain the same if the cellists produced a note with a higher frequency
Physics
1 answer:
sergij07 [2.7K]2 years ago
7 0

Answer:

Please see below as the answer is self-explanatory.

Explanation:

  • In order to have a destructive interference, the path difference between the sources of the sound, must be equal to an odd multiple of the semi-wavelength, as follows:
  • ⇒ d = d₂ - d₁ = n*(λ/2)
  • The minimum posible value for this distance, is when n=1, as it can be seen here:
  • dmin = λ/2
  • In any traveling wave, there exists a fixed relationship between the wave speed, the frequency and the wavelength:
  • v = λ*f
  • Therefore, assuming that the speed of sound keeps constant, if the frequency is increased, in order to keep the right side of the expression above balanced, λ must be decreased.
  • As the smallest separation that produces destructive interference is directly proportional to the wavelength, this means that this separation will decrease if the cellists produced a note with a higher frequency.

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Answer:

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A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
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Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

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speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

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Using eqn (2) and (3):

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