Question

Find the number of beats produced per sec by the vibrations ${x_1} = A\sin (320\pi t)$ and ${x_2} = A\sin (326\pi t)$ .
A) $3$
B) $4$
C) $5$
D) $6$

Answer 1

Hint: The number of beats produced by two different waves is the absolute difference between their frequencies. So, using the generation equation of wave will find angular frequency of the wave and using angular frequency of both waves, we will find frequencies of both waves and then absolute difference between found frequencies.

Complete step by step answer:
We know that the general equation of wave is given by,
$X = A\sin (\omega t)$
Where $X$ is the displacement of wave particle from it’s mean position
$A$ is the amplitude of the wave particle (maximum displacement of wave particle from it’s mean position)
$\omega $ (omega) is the angular frequency of wave and
$t$ is the time
Now, comparing this general equation with equation of first wave, we get,
$A\sin \left( {{\omega _1}t} \right) = A\sin \left( {320\pi t} \right)$
So, on comparing, we get,
${\omega _1} = 320\pi $
Now, angular frequency of wave $\left( \omega \right) = 2\pi f$
Where $f$ is frequency of wave,
So we get,
$2\pi {f_1} = 320\pi $
On solving we get,
${f_1} = 160$
Similarly comparing equation of second wave,
$A\sin \left( {{\omega _2}t} \right) = A\sin \left( {326\pi t} \right)$
On comparing we get,
$2\pi {f_2} = 326\pi $
On solving we get,
${f_2} = 163$
Now, beats produced by two different waves is difference between their frequencies,
\[{\text{beats = }}\left| {{f_2} - {f_1}} \right|\]
So we get,
\[{\text{beats = }}\left| {163 - 160} \right|\]
On solving, we get,
\[{\text{beats = 3}}\]

So the correct answer is option (A).

Note: It is important to note that beats are obtained by difference of frequencies and not angular frequencies, always convert angular frequency to frequency and only then find the difference in the frequency.