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The wave functions of two waves traveling in the same direction are given below. The two waves...

Question:

The wave functions of two waves traveling in the same direction are given below. The two waves have the same frequency, wavelength, and amplitude, but they differ in their phase constant.

{eq}y_1 (x,\ t) = 2 \sin (2 \pi x - 20 \pi t) {/eq}, and

{eq}y_2 (x,\ t) = 2 \sin (2 \pi x - 20 \pi t + \psi) {/eq},

where, {eq}y {/eq} is in centimetres, {eq}x {/eq} is in meters, and {eq}t {/eq} is in seconds.

The speed of the resultant wave due to the interference between {eq}y_1 {/eq} and {eq}y_2 {/eq} is:

a. 80 {eq}\pi {/eq} cm/s.

b. 10 m/s.

c. 20 m/s.

d. {eq}80 \pi \cos(\psi / 2) {/eq} cm/s.

e. 0.1 m/s.

f. Other.

The Wave Equation

All wave equations are derived from the same base formula,

{eq}y(x,t)=A \sin(kx-wt) {/eq}

In this equation, A is the amplitude, k represents wave number, w is frequency, and t is time.

When two waves travel on top of each other, they are added together. This is called superposition, and is defined by the equation

{eq}Y=y_{1}+y_{2} {/eq}

where Y is the sum of the two waves being added.

Answer and Explanation: 1

In order to solve the problem, the two given waves need to be substituted into the superposition equation.

{eq}y_1 (x,\ t) = 2 \sin (2 \pi x - 20 \pi t) {/eq}, and {eq}y_2 (x,\ t) = 2 \sin (2 \pi x - 20 \pi t + \psi) {/eq} will become

{eq}Y = 2 \sin (2 \pi x - 20 \pi t) + 2 \sin (2 \pi x - 20 \pi t + \psi) {/eq}

Using the trigonometric relation of

{eq}\sin(\alpha )+\sin(\beta)=2\sin\frac{1}{2}(\alpha +\beta )\cos\frac{1}{2}(\alpha -\beta ) {/eq}

the equation can be converted into

{eq}Y= 2(2\sin\frac{1}{2}((2 \pi x - 20 \pi t) +(2 \pi x - 20 \pi t + \psi) )\cos\frac{1}{2}((2 \pi x - 20 \pi t) -(2 \pi x - 20 \pi t + \psi) ) {/eq}

We can further simplify this expression to be:

{eq}Y=(2)(2)\sin(2 \pi x-20 \pi t+\frac{\psi}{2})\cos\frac{\psi}{2}) {/eq}

We can compare the simplified expression above to this general formula

{eq}Y=2A\sin(kx-wt+\frac{\psi}{2})\cos\frac{\psi}{2} {/eq}

to find the values we need to solve for the velocity of the combined wave.

Based on comparing these equations:

  • The amplitude A is 2cm.
  • Angular frequency ({eq}\omega {/eq}) is 20 {eq}\pi {/eq}, which can be divided by 2 {eq}\pi {/eq} to obtain a frequency ({eq}f {/eq}) of 10Hz.
  • {eq}k {/eq} is 2 {eq}\pi {/eq}. If 2 {eq}\pi {/eq} is divided by {eq}k {/eq}, it leaves the wavelength ({eq}\lambda {/eq}), which is 1cm in this case.

The speed of a wave is defined by the equation {eq}v=\lambda f {/eq}.

Plugging in our values for {eq}\lambda {/eq} and {eq}f {/eq}:

{eq}v=1(10)=10cm/s {/eq}.

After converting to meters, we see the correct answer is E) 0.1m/s


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