Question

Two sound waves (at $x=0$) are given by $y_1=4\sin \left(1026\pi t\right),y_2=2\sin \left(1014\pi t\right)$, where $y,t$ are in SI units. In case of superposition of these waves,

• A. the frequency of resulting wave is $510\text{Hz}$
• B. the amplitude of resulting wave varies at frequency $3\text{Hz}$
• C. the frequency of beats is $6\text{Hz}$.
• D. the maximum intensity of the resultant wave is 4 times the maximum intensity of each wave

Answer 1

Answer: (A), (B), (C)

The correct options are
A the frequency of resulting wave is $510\text{Hz}$
B the amplitude of resulting wave varies at frequency $3\text{Hz}$
C the frequency of beats is $6\text{Hz}$.

Number of beats heard
$=f_1-f_2=\frac{{\omega }_1}{2\pi }-\frac{{\omega }_2}{2\pi }$
$\Rightarrow \frac{1026\pi }{2\pi }-\frac{1014\pi }{2\pi }=6$
$c$ is correct

Maximum intensity $\propto (A_1+A_2{)}^2$
$\Rightarrow (4+2{)}^2=36$
Minimum intensity $\propto (A_1-A_2{)}^2$
$\Rightarrow (4-2{)}^2=4$
Ratio of maximum and minimum intensity
$\frac{I_{max}}{I_{min}}=\frac{36}{4}=\frac{9}{1}$
$d$ is incorrect.

$y_1=4\sin \left(1026\pi t\right)$
$y_2=2\sin \left(1014\pi t\right)$
$y=y_1+y_2=4\sin \left(1026\pi t\right)+2\sin \left(1014\pi t\right)$
So, clearly frequency
$=\frac{\omega }{2\pi }=\frac{1020\pi }{2\pi }\Rightarrow 510\text{Hz}$ and amplitude of resulting wave varies at frequency.
$\nu =\frac{\omega }{2\pi }=\frac{6\pi }{2\pi }=3\text{Hz}$