Ask Your Teacher Two wires, each of length 1.5 m, are stretched between two fixed supports. On wire A there is a second-harmonic standing wa

Question

Ask Your Teacher Two wires, each of length 1.5 m, are stretched between two fixed supports. On wire A there is a second-harmonic standing wave whose frequency is 456 Hz. However, the same frequency of 456 Hz is the third harmonic on wire B. Find the speed at which the individual waves travel on each wire.

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Trúc Chi 3 years 2021-08-12T21:11:42+00:00 2 Answers 27 views 0

Answers ( )

  1. thienhuong
    0
    2021-08-12T21:13:12+00:00
    Reply

    Answer:

    Explanation:

    Given:

    Length, L = 1.5 m

    Frequency, f = 456 Hz

    Velocity, v = frequency, f × wavelength, λ

    At the second harmonic,

    f = 2 × (v/2L)

    v = 456 × 1.5

    = 684 m/s

    At the third harmonic,

    f = 3 × (v/2L)

    Where,

    v = velocity

    v = 2/3 × 456 × 1.5

    = 456 m/s

  2. Dau
    0
    2021-08-12T21:13:17+00:00
    Reply

    Answer:

    On wire A the speed is 684 m/s

    On wire B the speed is 456 m/s

    Explanation:

    The velocity of a wave is defined as wavelength times frequency of the wave:

    v= \lambda f (1)

    with λ the wavelength and f the frequency of the wave.

    When you have a wire stretched between two fixed supports, you’re going to have standing waves if you perturbate the wire, and the wavelength of the different harmonics you can have is:

    \lambda_m=\frac{2L}{m}

    with L the length of the wire and m the number of the harmonic, so for the second harmonic:

    \lambda_2=\frac{2(1.5)}{2}=1.5m

    ad for the third harmonic:

    \lambda_3=\frac{2(1.5)}{3}=1.0m

    Using those values of wavelength on (1):

    For second harmonic:

    v_A=\lambda_2 f_A=1.5*456=684\frac{m}{s}

    For third harmonic:

    v_A=\lambda_3 f_A=1.0*456=456\frac{m}{s}

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