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azamat
3 years ago
7

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 .
Physics
1 answer:
NeX [460]3 years ago
4 0

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    

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Here,

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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
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71.4583 Hz

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The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

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\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

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Answer:

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Let's replace

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