Two loudspeakers in a plane, 5.0 m apart, are playing the same frequency. If you stand 12.0 m in front of the plane of the speak- ers, cente

Question

Two loudspeakers in a plane, 5.0 m apart, are playing the same frequency. If you stand 12.0 m in front of the plane of the speak- ers, centered between them, you hear a sound of maximum in- tensity. As you walk parallel to the plane of the speakers, staying 12.0 m in front of them, you first hear a minimum of sound inten- sity when you are directly in front of one of the speakers. What is the frequency of the sound? Assume a sound speed of 340 m/s

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Thanh Hà 6 months 2021-07-15T03:45:00+00:00 2 Answers 23 views 0

Answers ( )

    0
    2021-07-15T03:46:07+00:00

    Answer:

    The frequency of the sound is 170 Hz

    Explanation:

    Traveling wave is given as;

    D(r,t) = D_o(r)sin(kr- \omega t + \phi_o)

    where;

    r is the distance from the source

    (kr- \omega t + \phi_o) is phase

    Then phase difference is given as;

    \delta \phi = 2 \pi\frac{\delta r}{\lambda} +\delta \phi_o

    The phase difference between the two speakers at maximum intensity;

    \delta \phi = 2\pi \frac{\delta r}{\lambda}

    where;

    λ is the wavelength

    Δr is difference in distance between the two speakers

    Δr = r₂ – r₁ = \sqrt{L^2 + d^2} -L

    Given;

    distance between the two speakers, d = 5.0 m

    distance to the plane of the speakers, L = 12.0 m

    Δr = \sqrt{12^2 + 5^2} -12 = 1 \ m

    \delta \phi = 2\pi\frac{\delta r}{\lambda}

    \delta r = \frac{\delta \phi}{2\pi}\lambda, at minimum sound intensity, ΔФ = π

    \delta r = \frac{\pi}{2\pi} \lambda\\\\\delta r  = \frac{\lambda}{2}

    λ = 2Δr

    λ = 2 (1) = 2m

    f = \frac{v}{\lambda} = \frac{340}{2} = 170\ Hz

    Therefore, the frequency of the sound is 170 Hz

    0
    2021-07-15T03:46:23+00:00

    Answer:

    170 Hz

    Explanation:

    The path difference between two waves are given by:

    Δ=\sqrt{12^2 +5^2 } – 12

    Δ = 1 m

    In order to hear minimum intensity, the path difference will be

    Δ = λ/2

    λ = 2Δ

    Also, We know that

    V= f λ

    λ=V/f

    Therefore,

    340/f = 2 x 1

    f= 170 Hz

    Thus, the frequency of the sound is 170 Hz

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