# Two traveling waves of equal frequency, one of amplitude 4 cm and the other of amplitude 6 cm,...

## Question:

Two traveling waves of equal frequency, one of amplitude 4 cm and the other of amplitude 6 cm, superimpose in a single medium.

Which of the following best describes the displacement, D, of the resultant wave?

A. 2 cm {eq}\leq {/eq} D {eq}\leq {/eq} 10 cm

B. D = 5 cm

C. D = 10 cm

D. 10 cm {eq}\leq {/eq} D {eq}\leq {/eq} 12 cm

## Amplitude:

By using amplitude of a wave, the deviation is easily measured. The deviation is measured from the central value of the wave. By using the amplitude, the loudness of a wave is also obtained.

## Answer and Explanation:

**Given data**

- The amplitude of first wave is {eq}{a_1} = 4\;{\rm{cm}} {/eq}

- The amplitude of the second wave is {eq}{a_2} = 6\;{\rm{cm}} {/eq}

The expression of the displacement D, of the resultant wave for maximum superimposition is given by,

{eq}D = {a_1} + {a_2} {/eq}

Substitute all the values.

{eq}\begin{align*} D &= \left( {4\;{\rm{cm}}} \right) + \left( {6\;{\rm{cm}}} \right)\\ &= 10\;{\rm{cm}} \end{align*} {/eq}

The expression of the displacement D, of the resultant wave for minimum superimposition is given by,

{eq}D = {a_2} + {a_1} {/eq}

Substitute all the values.

{eq}\begin{align*} D &= \left( {6\;{\rm{cm}}} \right) - \left( {4\;{\rm{cm}}} \right)\\ &= 2\;{\rm{cm}} \end{align*} {/eq}

So, the range of the displacement is {eq}2\;{\rm{cm}} \le {\rm{D}} \le {\rm{10}}\;{\rm{cm}} {/eq}.

**Thus, the option (a) is correct.**

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from MEGA Physics: Practice & Study Guide

Chapter 15 / Lesson 10