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muminat
1 year ago
13

A sinusoidal sound wave moves through a medium and is described by the displacement wave function

Physics
1 answer:
labwork [276]1 year ago
7 0

By applying the wave equation we know that the displacement on the y-axis is 1.999 micrometers.

We need to know about wave equations to solve this problem. The displacement of the wave on the y-axis can be explained by the wave equation

y = A cos (kx - ωt)

where y is y-axis displacement, A is amplitude, k is wave number, x is x-axis displacement, ω is angular speed and t is time.

the wavenumber and angular speed of the wave equation can be determined respectively by

k = 2π / λ

ω = 2πf

where k is the wavenumber, λ is wavelength and f is frequency.

From the question above, we know that:

y = 2.00cos (15.7x - 858t)

(x in meters, t in second, y in micrometers)

x = 0.05 m

t = 3 ms

Convert time to second

t = 3ms = 0.003 s

By applying the wave equation, we get

y = 2.00cos (15.7x - 858t)

y = 2.00cos (15.7(0.05) - 858(0.003))

y  = 2 cos(-1.789)

y = 1.999 micrometers

For more on wave equation on: brainly.com/question/25699025

#SPJ4

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Plane polarized light with intensity I0 is incident on a polarizer. What angle should the principle axis make with respenct to t
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Answer:

 Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

                                     I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

                                     Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

                                     Q = cos^-1 ( sqrt (0.464))

                                     Q = cos^-1 (0.6811754546)

                                     Q = 47.06 degrees

                                   

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