Question

Two waves represented by $y_1=3\sin (2000\pi t-50x)$ and
$y_2=3\sin (2008\pi t+50x)$ are approaching each other. The number of beats heard per second will be
(Assume SI units)

• A. $8\text{Hz}$
• B. $4\text{Hz}$
• C. $1\text{Hz}$
• D. $5\text{Hz}$

Answer 1

The correct option is B $4\text{Hz}$

Given:
$y_1=3\sin (2000\pi t-50x)$
$y_2=3\sin (2008\pi t+50x)$
Frequency of the first wave,
$f_1=\frac{{\omega }_1}{2\pi }=\frac{2000\pi }{2\pi }=1000\text{Hz}$
and for the second wave,
$f_2=\frac{{\omega }_2}{2\pi }=\frac{2008\pi }{2\pi }=1004\text{Hz}$
Now, beat frequency,
$f_b=|f_1-f_2|=4\text{Hz}$
$\Rightarrow 4$ beats per second.