Question

Two waves are represented by $Y_1=4\sin 404\pi t$ and $Y_2=3\sin 400\pi t$. Then:

• A. Beat frequency is $4\text{Hz}$ and the ratio of maximum to minimum intensity is $49:1$
• B. Beat frequency is $2\text{Hz}$ and the ratio of maaximum to minimum intensity is $49:1$
• C. Beat frequency is $2\text{Hz}$ and the ratio of maximum to minimum intensity is $7:1$.
• D. Beat frequency is $4\text{Hz}$ and the ratio of maximum to minimum intensity is $7:1$

Answer 1

The correct option is B Beat frequency is $2\text{Hz}$ and the ratio of maaximum to minimum intensity is $49:1$

Given , $Y_1=4\sin 404\pi t,Y_2=3\sin 400\pi t$
Comparing with equation $Y=A\sin \omega t$,
${\omega }_1=404\pi ,{\omega }_2=400\pi ,A_1=4,A_2=3$

$\because {\omega }_1=2\pi f_1\Rightarrow 404\pi =2\pi f_1$
or $f_1=202\text{Hz}$

${\omega }_2=2\pi f_2\Rightarrow 400\pi =2\pi f_2$
or $f_2=200\text{Hz}$

Therefore,
Beat frequency $=f_1-f_2=202-200=2\text{Hz}$
and
Ratio of maximum and minimum intensity
$\frac{I_{max}}{I_{min}}={\left(\frac{A_1+A_2}{A_1-A_2}\right)}^2={\left(\frac{4+3}{4-3}\right)}^2$
$={\left(\frac{7}{1}\right)}^2=\frac{49}{1}$