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

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 8.25 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.540 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 = 1.51 m and t = 0.150 s .

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  1. Answer:

    The transverse displacement is   [tex]y(1.51 , 0.150) = 0.055 m[/tex]    

    Explanation:

     From the question we are told that

         The generally equation for the mechanical wave is

                        [tex]y(x,t) = Acos (kx -wt)[/tex]

         The speed of the transverse wave is [tex]v = 8.25 \ m/s[/tex]

         The amplitude of the transverse wave is [tex]A = 5.50 *10^{-2} m[/tex]

         The wavelength of the transverse wave is [tex]\lambda = 0540 m[/tex]

          At t= 0.150s , x = 1.51 m

     The angular frequency of the wave is mathematically represented as

              [tex]w = vk[/tex]

    Substituting values  

             [tex]w = 8.25 * 11.64[/tex]

            [tex]w = 96.03 \ rad/s[/tex]

    The propagation constant k is mathematically represented as

                      [tex]k = \frac{2 \pi}{\lambda}[/tex]

    Substituting values

                      [tex]k = \frac{2 * 3.142}{0. 540}[/tex]

                       [tex]k =11.64 m^{-1}[/tex]

    Substituting values into the equation for mechanical waves

          [tex]y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) – (96.03 * 0.150))[/tex]

           [tex]y(1.51 , 0.150) = 0.055 m[/tex]    

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