# Two waves traveling on a string in the same direction both have a frequency of 150 Hz, a...

## Question:

Two waves traveling on a string in the same direction both have a frequency of 150 Hz, a wavelength of 2 cm, and an amplitude of 0.06 m. What is the amplitude of the resultant wave if the original waves differ in phase by each of the following values?

a) {eq}\pi {/eq}/6

b) {eq}\pi {/eq}/3

## Amplitude:

The term amplitude can define as when one wave crest or trough overlaps or coincides with the other wave and to form a resultant wave. In S.I system, its measurable units are meter.

Given data:

• The frequency is {eq}f = 150\,{\rm{Hz}} {/eq}
• The amplitude of each wave is {eq}{A_1} = {A_2} = 0.06\,{\rm{m}} = 6\,{\rm{cm}} {/eq}

(a)

The expression for the resultant wave amplitude is

{eq}A = \sqrt {{A_1}^2 + {A_2}^2 + 2{A_1}{A_2}\cos \phi } {/eq}

• Here {eq}\phi = \dfrac{\pi }{6} {/eq} is the phase angle.

Substituting the values in the above equation as,

{eq}\begin{align*} A &= \sqrt {{A_1}^2 + {A_2}^2 + 2{A_1}{A_2}\cos \phi } \\ A &= \sqrt {{{\left( 6 \right)}^2} + {{\left( 6 \right)}^2} + 2\left( 6 \right)\left( 6 \right)\cos \dfrac{\pi }{6}} \\ A &= \sqrt {36 + 36 + 72 \times 0.866} \\ A &= \sqrt {134.352} \\ A &= 11.59\,{\rm{cm}} \end{align*} {/eq}

Thus the amplitude is {eq}A = 11.59\,{\rm{cm}} {/eq}

(b)

As from the given data, the phase angle is {eq}\phi = \dfrac{\pi }{3} {/eq}

The expression for the resultant wave amplitude is

{eq}A = \sqrt {{A_1}^2 + {A_2}^2 + 2{A_1}{A_2}\cos \phi } {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} A &= \sqrt {{A_1}^2 + {A_2}^2 + 2{A_1}{A_2}\cos \phi } \\ A &= \sqrt {{{\left( 6 \right)}^2} + {{\left( 6 \right)}^2} + 2\left( 6 \right)\left( 6 \right)\cos \dfrac{\pi }{3}} \\ A &= \sqrt {36 + 36 + 72 \times 0.5} \\ A &= 10.39\,{\rm{cm}} \end{align*} {/eq}

Thus the amplitude of the resultant wave is {eq}A = 10.39\,{\rm{cm}} {/eq}