A transverse wave is traveling on a string. The displacement y of a particle from its equilibrium position is given by y = (0.021 m)sin(29t

Question

A transverse wave is traveling on a string. The displacement y of a particle from its equilibrium position is given by y = (0.021 m)sin(29t – 2x). Note that the phase angle (29t – 2.0x) is in radians, t is in seconds, and x is in meters. The linear density of the string is 1.80 10-2 kg/m. What is the tension in the string?

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Verity 4 years 2021-08-09T06:50:20+00:00 1 Answers 12 views 0

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  1. thucuc
    0
    2021-08-09T06:52:02+00:00
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    Answer: 3.7845 N

    Explanation: The wave equation is given by

    y = 0.021 sin (29t – 2x)

    By comparing this to the general equation of a wave, we have that

    y = A sin (ωt – kx)

    Where ω = angular frequency.

    After comparing, we see that

    A = 0.021m, ωt = 29t and – 2x = – kx

    Hence ω = 29 rad/s and k = 2.

    The velocity of a wave is in relationship with angular frequency and wave number is given as

    ω = kv

    Where k = wave number

    By substituting the parameters, we have that

    29 = 2v

    v = 29/2 = 14.5 m/s.

    The velocity of wave in a spring is calculated using the formulae below.

    v = √(T/u)

    Where v = linear velocity = 14.5 m/s

    T = tension =?

    u = linear density = 1.80×10^-2 kg/m

    By substituting the parameters, we have that

    14.5 = √(T/ 1.80×10^-2)

    By squaring both sides, we have that

    14.5² = T/ 1.80×10^-2

    T = 14.5² × 1.80×10^-2

    T = 210.25 × 1.80×10^-2

    T = 3.7845 N

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