Transverse waves on a string have wave speed v=8.00 m/s, amplitude A=0.0700m, and direction, and at t=0 the x-0 end of the wavelength -0.320

Question

Transverse waves on a string have wave speed v=8.00 m/s, amplitude A=0.0700m, and direction, and at t=0 the x-0 end of the wavelength -0.320m. The waves travel in the -x string has its maximum upward displacement. a) Find the frequency, period and wave number of these waves b) Write a wave function describing the wave c) Find the transverse displacement of a particle at x=0.360m at time t=0.150 d) How much time must elapse from the instant in part (c) until the particle at x-0.360 m next has maximum upward displacement?

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Jezebel 4 years 2021-08-23T02:41:46+00:00 1 Answers 24 views 0

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  1. Orla
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    2021-08-23T02:43:45+00:00
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    Answer:

    a. frequency = 25 Hz, period = 0.04 s , wave number = 19.63 rad/m

    b. y = (0.0700 m)sin[(19.63 rad/m)x – (157.08 rad/s)t]

    c.  0.0496 m

    d. 0.03 s

    Explanation:

    a. Frequency, f = v/λ where v = wave speed = 8.00 m/s and λ = wavelength = 0.320 m

    f = v/λ = 8.00 m/s ÷ 0.320 m = 25 Hz

    Period, T = 1/f = 1/25 = 0.04 s

    Wave number k = 2π/λ = 2π/0.320 m = 19.63 rad-m⁻¹

    b. Using y = Asin(kx – ωt) the equation of a wave

    where y = displacement of the wave, A = amplitude of wave = 0.0700 m and ω = angular speed of wave = 2π/T = 2π/0.04 s = 157.08 rad/s

    Substituting the variables into y, we have

    y = (0.0700 m)sin[(19.63 rad/m)x – (157.08 rad/s)t]

    c. When x = 0.360 m and t = 0.150 s, we substitute these into y to obtain

    y = (0.0700 m)sin[(19.63 rad/m)x – (157.08 rad/s)t]

    y = (0.0700 m)sin[(19.63 rad/m × 0.360 m) – (157.08 rad/s × 0.150 s)]

    y = (0.0700 m)sin[(7.0668 rad) – (23.562 rad)]

    y = (0.0700 m)sin[-16.4952 rad]

    y = (0.0700 m) × 0.7084

    y = 0.0496 m

    d. For the particle at x = 0.360 m to reach its next maximum displacement, y = 0.0700 m at time t. So,

    y = (0.0700 m)sin[(19.63 rad/m)x – (157.08 rad/s)t]

    0.0700 m = (0.0700 m)sin[(19.63 rad/m × 0.360 m) – (157.08 rad/s)t]

    0.0700 m = (0.0700 m)sin[(7.0668 rad – (157.08 rad/s)t]  

    Dividing through by 0.0700 m, we have

    1 = sin[(7.0668 rad – (157.08 rad/s)t]

    sin⁻¹(1) = 7.0668 rad – (157.08 rad/s)t  

    π/2 = 7.0668 rad – (157.08 rad/s)t

    π/2 – 7.0668 rad = – (157.08 rad/s)t

    -5.496 rad = – (157.08 rad/s)t

    t = -5.496 rad/(-157.08 rad/s) = 0.03 s

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