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Find the sum y of the following quantities: y1=10sinӬt and y2=8.0sin(Ӭt+30°)

Short Answer

Expert verified

The sum of wave is 17sinӬt+13.3°.

Step by step solution

01

Identification of given data

The amplitude of first wave is A1=10

The amplitude of second wave is A2=8

The phase difference for both waves is ϕ=30°

The amplitude of the resultant wave is equal to the vector sum of the amplitude of each wave.

02

Determination of resultant amplitude and direction of resultant amplitude of wave

The resultant amplitude of wave is given as:

A=A12+A22+2A1A2cosϕ

Substitute all the values in equation.

A=102+82+2108cos30°A=302.56A≈17

The direction of the resultant amplitude is given as:

role="math" localid="1663050213428" tanθ=A2sinϕA1+A2cosϕ

Substitute all the values in equation.

tanθ=8sin30°10+8cos30°θ=13.3°

03

Determination of sum of wave

The sum of the wave is given as:

y=AsinӬt+θ

Substitute all the values in equation.

y=17sinӬt+13.3°

Therefore, the sum of wave is 17sinӬt+13.3°.

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Most popular questions from this chapter

Transmission through thin layers. In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of r3and r4interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1,n2and n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength λin nanometers of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.

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Transmission through thin layers. In Fig. 35-43, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3 (the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of r3 and r4 interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1, n2, and n3, the type of interference, the thin-layer thickness L in nanometers, and the wavelength λ in nanometers of the light as measured in air. Where λ is missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.

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