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Learning Goal: To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 9.00 m/s , an amplitude of 9.00×10−2 m , and a wavelength of 0.480 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.59 m and t = 0.150 s .

Respuesta :

Answer:

so the transverse displacement is  0.089963 m

Explanation:

Given data

equation y(x,t)=Acos(kx−ωt)

speed  v = 9.00 m/s

amplitude A = 9.00 × 10^−2 m

wavelength λ   = 0.480 m

to find out

the transverse displacement

solution

we know

v = angular frequency / wave number

and

wave number = 2 [tex]\pi[/tex] / λ  =  2 [tex]\pi[/tex] / 0.480  = 13.0899 [tex]m^{-2}[/tex]

angular frequency = v k

angular frequency = 9.00 × 13.0899

angular frequency = 117.8097 rad/sec = 118 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)

when x =0 and and t = 0

maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))

maximum y(x,t)= 9.00 × 10^−2  m

and when x =  x = 1.59 m and t = 0.150 s

y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )

y(x,t)=9.00 × 10^−2 × (0.99959)

y(x,t) = 0.089963 m

so the transverse displacement is  0.089963 m

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