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23 February, 11:54

A traveling wave on a string can be described by the equation : y = (5.26 ~/text{m}) / cdot / sin / big ((1.65 ~/frac{/text{rad}}{/text{m}}) x - (4.64 ~/frac{/text{rad}}{/text{sec}}) t + (1.33 ~/text{rad}) / big) y = (5.26 m) ⋅sin ((1.65 m rad ) x - (4.64 sec rad ) t + (1.33 rad)) How much time will it take for a peak on this traveling wave to propagate a distance of 5.00 meters along the length of the string?

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  1. 23 February, 14:21
    0
    t = 1.77 s

    Explanation:

    The equation of a traveling wave is

    y = A sin [2π (x / λ - t / T) ]

    where A is the oscillation amplitude, λ the wavelength and T the period

    the speed of the wave is constant and is given by

    v = λ f

    Where the frequency and period are related

    f = 1 / T

    we substitute

    v = λ / T

    let's develop the initial equation

    y = A sin [ (2π / λ) x - (2π / T) t + Ф]

    where Ф is a phase constant given by the initial conditions

    the equation given in the problem is

    y = 5.26 sin (1.65 x - 4.64 t + 1.33)

    if we compare the terms of the two equations

    2π / λ = 1.65

    λ = 2π / 1.65

    λ = 3.81 m

    2π / T = 4.64

    T = 2π / 4.64

    T = 1.35 s

    we seek the speed of the wave

    v = 3.81 / 1.35

    v = 2.82 m / s

    Since this speed is constant, we use the uniformly moving ratios

    v = d / t

    t = d / v

    t = 5 / 2.82

    t = 1.77 s
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