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    2 Problems on Wave Motion: Doppler effect, sound waves

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    (See attached file for full problem description)

    Please, on problem 3, a bat is flying...., where does the number 345 come from and why used?

    Problem 4, explain how parts A,B, and C were solved, where equations come from, etc.

    See attached below, next two pages.

    © BrainMass Inc. brainmass.com December 24, 2021, 5:42 pm ad1c9bdddf
    https://brainmass.com/physics/doppler-shift/problems-wave-motion-doppler-effect-sound-waves-63577

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    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see the attachment.

    Please, on problem 3, a bat is flying...., where does the number 345 come from and why used?

    Answer
    This formula is from Doppler effect. The apparent frequency of sound wave when the source and listener are moving relatively is given by this formula, here 345 is the velocity of sound in air at normal temperature. In some problems of sound speed of sound is not given, in such cases we take common value of speed of sound in air = 345 m/s (sometimes 340m/s also).

    The formula for apparent frequency is
    Where c is the speed of wave in the medium and the signs are according to direction of motion of source and listener. ( please go through the text).

    Problem 4, explain how parts A,B, and C were solved, where equations come from, etc.

    See attached below, next two pages.

    Answer:

    The first graph is showing the variation in displacement y of particles of medium at a particular instant of time (t = constant) with distance from the source x. This shows us that the situation y is repeated after distance x = 3 (m). it means that two particles 3 m away are undergoing same wave displacement y and hence the wavelength  = 3m. As well as the maximum displacement of a particle from equilibrium position is 2 mm and this is the amplitude ym of the wave.

    The second graph shows the variation of displacement y of a particular particle (x = constant) with time. Hear we see that in 10 ms (10*10-3s) the particle is making 1.5 oscillation exactly and hence the time period of the wave is T = 10/1.5 = 6.667 per ms.

    Now as the speed of the wave is given by v = f  = (1/T)  and solving we get the velocity, here f is the frequency of the wave.

    K is called wave number the number of ripples in unit length and given by 2/ = 2/3 and  is the wave frequency = 2/T = 2/(6.667*10-3) = 300rad/s.

    As from the graphs at t= o and x = 0; y = 0 the initial phase of the wave  is zero. The direction of wave motion is not given hence for the term t we have taken both + and - signs and substituting the values the equation can be written in the given form as

    differentiating above equation with respect to time we get speed of a particle at time t as
    c = dy/dt = (2mm)*300*cos
    and hence the maximum speed of the particle is given by (2mm)*300because the maximum value of cosine function is one.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:42 pm ad1c9bdddf>
    https://brainmass.com/physics/doppler-shift/problems-wave-motion-doppler-effect-sound-waves-63577

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