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The equation of a transverse wave traveling along a very long string is y = 6.0 sin(0.020πx + 4.0πt), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at x = 3.5 cm when t = 0.26 s?

Answer :

Answer:

a) A = 6.0 cm , b)  λ = 100 cm , c)  f = 0.5 Hz, d) v = 0.5 m / s, e)  the wave propagates to the right , f)  v = 0.12π m / s, g)   y = -1.67 cm

Explanation:

The general equation of a wave on a string going off to the right is

          y = A sin (kx - wt)

where A is the amplitude of the movement. w the angular velocity and k the wave number

in this exercise the equation remain is

        y = 6.0 sim (0.020π x - 4.0π t)

a) Let's compare the two equations to find the answers

       the amplitude is

         A = 6.0 cm

b) The wave number is

         k = 2π/λ

         λ = 2π / k

         λ = 2π / 0.020π

         λ = 100 cm

c) angular velocity is related to frequency

          w = 2π f

          f = 2π / w

          f = 2π / 4.0π

          f = 0.5 Hz

d) at wave speed is given by

          v = λ f

          v = 1   0.5

          v = 0.5 m / s

e) as the temporal part is negative, the wave propagates to the right

f) we find the maximum transverse speed from the equation

       v = dy / dt

       v = -A w cos kx-wt)

the speed is maximum when the cosine function is ±1

        v = -A w

         v = 6.0  0.020π

         v = 0.12π m / s

g) the displacement for x = 3.5 cm t = 0.26 s

          y = 6.0 sin (0.020π 6.5 - 4.0π 0.26)

          y = 6.0 (-0.279)

          y = -1.67 cm

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