Question 129: Waves

Question

Two harmonic waves moving in the same direction superimpose to form a wave $x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t})$ where t is in seconds. Find the period with which they beat. (close to nearest integer)

Accepted Answer

The superposition of the two harmonic waves results in the wave equation:
$  x = a \cos(1.5t) \cos(50.5t)  $
This equation can be expanded using the trigonometric identity for the product of cosines:
$ x = \frac{a}{2} \cos[(1.5 + 50.5)t] + \frac{a}{2} \cos[(50.5 - 1.5)t] $
Simplifying, we find:
$ x = \frac{a}{2} \cos(52t) + \frac{a}{2} \cos(49t) $
This represents two waves with angular frequencies $52$ and $49$, respectively.
To find the beat frequency, we calculate the differences in frequencies:
$ f_1 = \frac{52}{2\pi}, \quad f_2 = \frac{49}{2\pi} $
The beat frequency is then:
$ f_{	ext{Beat}} = f_1 - f_2 = \frac{3}{2\pi} 	ext{ Hz} $
The period of the beats, $T_{	ext{Beat}}$, is the reciprocal of the beat frequency:
$ T_{	ext{Beat}} = \frac{1}{f_{	ext{Beat}}} = \frac{2\pi}{3} 	ext{ sec} $
Approximating this gives:
$ T_{	ext{Beat}} \approx 2.09 	ext{ sec} \approx 2 	ext{ sec} $

Answer

1 s

Answer

4 s

Answer

2 s

Answer

6 s

Waves

Two harmonic waves moving in the same direction superimpose to form a wave x=\mathrm{a} \cos (1.5 \mathrm{t}) \cos (50.5 \mathrm{t}) where t is in seconds. Find the period with which they beat. (close to nearest integer)

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