A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s

Respuesta :

Answer:

a)  A = 1.99 μm , b) λ = 0.4134 m , c)  v = 57.2 m / s , d)   s = - 1,946 nm ,

e)      v_max = 1,739 mm / s

Explanation:

A sound wave has the general expression

           s = s₀ sin (kx - wt)

where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is

           s = 1.99 sin (15.2 x - 869 t)

a) the amplitude of the wave is

        A = s₀

        A = 1.99 μm

b) wave spectrum is

      k = 2π /λ

in the equation k = 15.2 m⁻¹

      λ = 2π / k

      λ = 2π / 15.2

     λ = 0.4134 m

c) the speed of the wave is given by the relation

       v = λ f

angular velocity and frequency are related

       w = 2π f

        f = w / 2π

        f = 869 / 2π

        f = 138.3 Hz

   

        v = 0.4134 138.3

         v = 57.2 m / s

d) To find the instantaneous velocity, we substitute the given distance and time into the equation

       s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

       s = 1.99 sin (0.77368 - 2.55486)

remember that trigonometry functions must be in radians

       s = 1.99 (-0.98895)

       s = - 1,946 nm

The negative sign indicates that it shifts to the left

e) the speed of the oscillating part is

           v = ds / dt)

           v = - s₀(-w) cos (kx -wt)

the maximum speed occurs when the cosines is 1

           v_maximo = s₀w

           v_maximum = 1.99 869

           v_maximo = 1739.31 μm / s

let's reduce to mm / s

          v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

          v_max = 1,739 mm / s

a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s

Calculation of Wavelength

When A sound wave has the general expression is:

Then, s = s₀ sin (kx - wt)

Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is

s is = 1.99 sin (15.2 x - 869 t)

a) When the amplitude of the wave is

A is = s₀

Thus, A = 1.99 μm

b) When the wave spectrum is

k is = 2π /λ

Now, in the equation k = 15.2 m⁻¹

Then, λ = 2π / k

After that, λ = 2π / 15.2

Thus, λ = 0.4134 m

c) When the speed of the wave is given by the relation is:

Then, v = λ f

Now, the angular velocity and frequency are related is:

w is = 2π f

Then, f = w / 2π

After that, f = 869 / 2π

Now, f = 138.3 Hz

Then, v = 0.4134 138.3

Thus, v = 57.2 m / s

d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation

Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)

After that, s = 1.99 sin (0.77368 - 2.55486)

Then remember that trigonometry functions must be in radians

After that, s = 1.99 (-0.98895)

Thus, s = - 1,946 nm

When The negative sign indicates that it shifts to the left

e) When the speed of the oscillating part is

Then, v = ds / dt)

Now, v = - s₀(-w) cos (kx -wt)

When the maximum speed occurs when the cosines is 1

Then, v_maximo = s₀w

After that, v_maximum = 1.99 869

v_maximo = 1739.31 μm / s

Now, let's reduce to mm / s

Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)

Therefore, v_max = 1,739 mm / s

Finf more informmation about Wavelength here:

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