Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in

Question

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

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Đan Thu 6 months 2021-07-18T01:02:07+00:00 1 Answers 25 views 0

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    2021-07-18T01:04:01+00:00

    Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

    Explanation:

    It is known that formula for path difference is as follows.

    \Delta L = (n + \frac{1}{2}) \times \frac{\lambda}{2}    … (1)

    where, n = 0, 1, 2, and so on

    As John is standing perpendicular to the line joining the speakers. So, the value of L_{1} is calculated as follows.

    L_{1} = \sqrt{(2)^{2} + (5)^{2}}\\= 5.4 m

    Hence, path difference is as follows.

    \Delta L = (5.4 - 5) m = 0.4 m

    For lowest frequency, the value of n = 0.

    \Delta L = (0 + \frac{1}{2}) \times \frac{\lambda}{2} = \frac{\lambda}{4}

    \lambda = 4 \Delta L

    where,

    \lambda = wavelength

    The relation between wavelength, speed and frequency is as follows.

    \lambda = \frac{\nu}{f}\\4 \Delta L = \frac{\nu}{f}\\

    where,

    \nu = speed

    f = frequency

    Substitute the values into above formula as follows.

    f = \frac{\nu}{4 \Delta L}\\f = \frac{340}{4 \times 0.4 m}\\= 212.5 Hz

    Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.

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