Answer:
Explanation:
ocities: 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 itNBself).Consider a transverse wave traveling in a string. The mathematical form of the wave is: y(x,t) = A sin(kx - ωt)Part AFind the velocity of propagation v_p of this wave.Express the velocity of propagation in terms ofNGHJGHHG some or all of the variables A, k, and ω.Part BFind the y velocity v_y(x,t) of a point on the string as a function of x and t.Express the y velocity in terms of ω, A, k, x, and t.Part CWhich of the following statements about v_x(x,t), the x component of the velocity of the string, is true?A) v_x(x;t) = v_pB) v_x(x;t) = v_y(x;t)C) v_x(x;t) has the same mathematical form as v_y(x;t) but is 180° out of phase.D) v_x(x;t)=0Part DFind the slope of the string ∂_y(x,t) / ∂_x as a function of position x and time t.Express your answer in terms of A,k, ω, x, and t.NNNNN
a) The velocity of propogation of the wave V=w/k
b) The y velocity v_y(x,t) of a point on the string as a function of x v=-wAcos(kx-wt)
A wave can be described as a disturbance that travels through a medium from one location to another location
y(x,t)=Asin(kx−ωt) defines the wave equation.
a)The velocity of propogation of the wave
We are asked to find wave speed (v)
Recall that v = fλ
From the wave equation above,
k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k
ω = 2πf where f is the frequency and ω is the angular frequency.
f = ω/ 2π.
By substituting for λ and ω into the wave speed formulae, we have that
v =( ω/ 2π) × (2π /k)
v = ω/k
b)The y velocity v_y(x,t) of a point on the string as a function of x
y(x,t)=Asin(kx−ωt)
The first derivative of y with respect to x give the velocity (vy)
By using chain rule, we have that
v = dy/dt = A cos( kx −ωt) × (−ω)
v = - ωAcos( kx −ωt)
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