Question

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 .

Answer 1

Answer:

The transverse displacement is   $y(1.51 , 0.150) = 0.055 m$    

Explanation:

 From the question we are told that

     The generally equation for the mechanical wave is

                    $y(x,t) = Acos (kx -wt)$

     The speed of the transverse wave is $v = 8.25 \ m/s$

     The amplitude of the transverse wave is $A = 5.50 *10^{-2} m$

     The wavelength of the transverse wave is $\lambda = 0540 m$

      At t= 0.150s , x = 1.51 m

 The angular frequency of the wave is mathematically represented as

          $w = vk$

Substituting values  

         $w = 8.25 * 11.64$

        $w = 96.03 \ rad/s$

The propagation constant k is mathematically represented as

                  $k = \frac{2 \pi}{\lambda}$

Substituting values

                  $k = \frac{2 * 3.142}{0. 540}$

                   $k =11.64 m^{-1}$

Substituting values into the equation for mechanical waves

      $y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))$

       $y(1.51 , 0.150) = 0.055 m$