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Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself). Consider a transverse wave traveling in a string. The mathematical form of the wave is y(x,t)=Asin(kx−ωt). a. Find the speed of propagation vp of this wave.b. Find the y velocity vy(x,t) of a point on the string as a function of x and t.

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Answer with Explanation:

We are given that  the mathematical form of wave

[tex]y(x,t)=Asin(kx-\omega t)[/tex]

a.We have to find the speed of propagation of this wave

In given mathematical form

k=Wave number

[tex]\omega[/tex]=Angular frequency

A=Amplitude

We know that

Speed of propagation of the wave=[tex]v_p=\frac{\omega}{k}[/tex]

b.Differentiate w.r.r t

[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]

The velocity [tex]v_y(x,t)[/tex] of a point on the string as  a function of x and t is given  by

[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]

onyebuchinnaji

(a) The speed of propagation of the wave is [tex]\frac{\omega}{k}[/tex]

(b) (b) The velocity of a point on the string as a function of x and t is [tex]v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]

The given parameters;

[tex]y(x, t) = Asin(kx - \omega t)[/tex]

where;

  • t is the time of motion
  • y is the displacement
  • k is the wave number
  • ω is the angular frequency
  • A  is the amplitude of the wave

(a) The speed of propagation of the wave is calculated as follows;

[tex]v_p = \frac{\omega }{k}[/tex]

(b) The velocity of a point on the string as a function of x and t is calculated as follows;

[tex]v = \frac{dy}{dt} \\\\v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]

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