Question

Two waves whose equations of displacement of particles in the medium are $y_1=4\sin (500\pi t-25x)$ and $y_2=2\sin (506\pi t-30x)$. How many beats will generate due to the superposition of the two waves and what will be the relation between maximum $\left(I_{max}\right)$ and minimum $\left(I_{min}\right)$ intensity?

• A. $3$ beats per second and $I_{max}=2I_{\text{min}}$
• B. $3$ beats per second and $I_{max}=9I_{\text{min}}$
• C. $6$ beats per second and $I_{max}=2I_{\text{min}}$
• D. $6$ beats per second and $I_{max}=9I_{\text{min}}$

Answer 1

The correct option is B $3$ beats per second and $I_{max}=9I_{\text{min}}$

Given:
$y_1=4\sin (500\pi t-25x)$
$y_2=2\sin (506\pi t-30x)$
Frequency of the first wave,
$f_1=\frac{{\omega }_1}{2\pi }=\frac{500\pi }{2\pi }=250\text{Hz}$
and for the second wave,
$f_2=\frac{{\omega }_2}{2\pi }=\frac{506\pi }{2\pi }=253\text{Hz}$
Now, beat frequency,
$f_b=|f_1-f_2|=3\text{Hz}$
$\Rightarrow 3$ beats per second.
Now,
$\frac{I_{max}}{I_{min}}={\left(\frac{A_{max}}{A_{min}}\right)}^2$
$\Rightarrow \frac{I_{\text{max}}}{I_{\text{min}}}={\left(\frac{A_1+A_2}{A_1-A_2}\right)}^2={\left(\frac{4+2}{4-2}\right)}^2$
$\Rightarrow \frac{I_{\text{max}}}{I_{\text{min}}}=\frac{9}{1}$
$\Rightarrow I_{\text{max}}=9I_{\text{min}}$