Question

Two travelling sinusoidal waves are described by the wave functions $y_1=10\sin [\pi (8x-1400t)]$
and $y_2=10\sin [\pi (8x-1400t-0.5)]$ where $x$, $y_1$, and $y_2$ are in metres and $t$ is in seconds. Which of the following option(s) is/are correct ?

• A. Amplitude of the resultant wave is $10\sqrt{2}m$
• B. Frequency of the resultant wave is $700$ $Hz$
• C. Time period of the resultant wave is $0.1$ $s$
• D. None of these

Answer 1

Answer: (A), (B)

Frequency of the resultant wave is $700$ $Hz$

Given wave functions are
$y_1=10\sin [\pi (8x-1400t)].....\left(1\right)$
$y_2=10\sin [\pi (8x-1400t-0.5)]......\left(2\right)$
Comparing these with the general equation,
$y=A\sin (kx-\omega t+\varphi )$
Phase difference between $\left(1\right)$ and $\left(2\right)$ is given by $\varphi =\frac{\pi }{2}$

Now, according to the principle of superposition, the resultant wave function will be :
$\overrightarrow{y}={\overrightarrow{y}}_1+{\overrightarrow{y}}_2$
From $\left(1\right)$ and $\left(2\right)$, we can conclude that the waves are moving in the same direction, so we can write the above equation as
$y_r=y_1+y_2$
$\Rightarrow y_r=2A_0\cos \left(\frac{\varphi }{2}\right)\sin (kx-\omega t-\frac{\varphi }{2})$
Using the given data,
$y_r=(2\times 10)\cos \left(\frac{\pi }{4}\right)\sin [\pi (8x-1400t)-\frac{\pi }{4}]$
$\Rightarrow y_r=10\sqrt{2}\sin [\pi (8x-1400t)-\frac{\pi }{4}]$

So, Resultant amplitude $A_r=10\sqrt{2}\text{m}$
Resultant Angular frequency ${\omega }_r=1400\pi$
Frequency of resultant wave $f=\frac{1400\pi }{2\pi }=700\text{Hz}$
$\therefore$ Time period $\left(T\right)=\frac{1}{f}=\frac{1}{700}=0.0014\text{sec}$

Thus, options (a) and (b) are the correct answers.