Question

When beats are formed by two waves of frequencies n${}_1$ and n${}_2$, the amplitude varies with frequency equal to

• A. $n_1-n_2$
• B. $2(n_1-n_2)$
• C. $(n_1-n_2)/2$
• D. $(n_1+n_2)/2$

Answer 1

The correct option is C $(n_1-n_2)/2$

Beats are formed due to the superposition/interference of waves

$y_2=Asin{\omega }_{1}t$

$y_2=Asin{\omega }_{2}t$

$y_1+y_2=A(sin{\omega }_1+sin{\omega }_{2}t)$

$2Asin\frac{({\omega }_1+{\omega }_2)}{2}tcos\left(\frac{{\omega }_1-{\omega }_2}{2}\right)t$

$y=A^{1}sin{\omega }^{1}t$

$w^1=\left(\frac{{\omega }_1+{\omega }_2}{2}\right)$

$A^1=2Acos\left(\frac{{\omega }_1-{\omega }_2}{2}\right)t$

We know that, frequency (n) is directly proportional to the angular frequency

($\omega$).

Here, the amplitude varies as $\left(\frac{{\omega }_1-{\omega }_2}{2}\right)$.

So, the amplitude varies as $\frac{n_1-n_2}{2}$, where $n_1$ and $n_2$ are frequencies.