# Two waves traveling on a string in the same direction both have a frequency of 92 Hz, a...

## Question:

Two waves traveling on a string in the same direction both have a frequency of 92 Hz, a wavelength of 0.19 m, and an amplitude of 0.32 m. What is the amplitude of the resultant wave if the original waves differ in phase by {eq}\frac{\pi }{3}{/eq} rad? What is the phase difference between the two waves if the amplitude of the resultant wave is 0.25 m?

## Interference of Waves

When two waves of same frequency *F*, same wavelength {eq}\lambda {/eq}, and having amplitude *A* of the form {eq}y_1 = A \sin ( \dfrac{ 2 \pi } { \lambda } x - \dfrac {2 \pi } { T } t ) {/eq} and {eq}y_2 = A \sin ( \dfrac { 2\pi } { \lambda } x - \dfrac{ 2 \pi } { T } t + \phi ) {/eq} travel in the same direction superimpose each other and create constructive and destructive interference. The term {eq}\phi {/eq} is the phase difference between the two waves. The resultant wave is of the form {eq}y = 2 A \cos ( \dfrac { \phi } { 2 } ) \sin ( \dfrac { 2 \pi } { \lambda } x - \dfrac { 2\pi } { T } t + \dfrac { \phi } { 2 } ) {/eq}. In this equation the **cos** part is the amplitude part and if the phase difference is zero we get the maximum amplitude as twice the amplitude of the interfering waves and minimum as of zero amplitude. The **sin** part is the phase part. When this part is maximum we get maximum intensity and when this part becomes minimum we get minimum intensity.

## Answer and Explanation:

**Given data**

- Frequency of both the waves {eq}F = 92 \ Hz {/eq}

- Wavelength of both the waves {eq}\lambda = 0.19 \ m {/eq}

- Amplitude of each wave {eq}A = 0.32 \ m {/eq}

- Phase difference between the waves {eq}\phi = \dfrac { \pi } { 3 } \ rad {/eq}

**Part a )**

Angular frequency of the wave {eq}\omega = 2 \pi F \\ \omega = 2 \pi \times 92 \\ \omega = 184 \pi \ rad {/eq}

Wave number of the wave {eq}k = \dfrac { 2 \pi } { \lambda } \\ k = \dfrac { 2 \pi } { 0.19 } \\ k = 10.53 \pi \ / m {/eq}

Then the individual waves can be expressed as {eq}y_1 = 0.32 \sin ( 10.53 \pi x - 184 \pi t ) {/eq}

and the second wave {eq}y_2 = 0.32 \sin ( 10.53 \pi x - 184 \pi t + \dfrac { \phi } { 3 } ) {/eq}

The wave resulting from the superposition of the two can be written as {eq}y = 2\times 0.32 \cos ( \dfrac { \pi } { 6 } ) \sin ( 10.53 \pi x - 184 \pi t + \dfrac { \pi } { 6 } ) {/eq}

The maximum amplitude occurs when the phase part becomes equal to one.

The maximum amplitude {eq}y_m = 2 \times 0.32 \times \cos ( \dfrac { \pi } { 6 } ) \\ y_m = 0.554 \ m {/eq}

**Part b)**

It is given that the resultant amplitude {eq}y_{m1} = 0.25 \ m {/eq}

Let {eq}\phi_1 {/eq} be the phase difference between the waves.

Then the amplitude of the resultant wave can be expressed as {eq}y_{m1} = 2 \times A \cos ( \dfrac { \phi_1 } { 2 } ) \\ \cos ( \dfrac{ \phi_1 } { 2 } ) = \dfrac { y_{m1} } { 2 A } {/eq}

Therefore the phase difference between the waves {eq}\phi_1 = 2 \times \cos^{-1} ( \dfrac { y_{m1} } { 2 A } ) \\ \phi_1 = 2 \times \cos ^{-1} ( \dfrac { 0.25 } { 2 \times 0.32 } ) \\ \phi_1 = 2.34 \ rad \\ \phi_1 = 134 ^o {/eq}

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 8 / Lesson 16