Question

Find beat frequency. Motion of two particles is given by $y_1=0.25\sin \left(310t\right)$ and $y_2=0.25\sin \left(316t\right)$

• A. $3$
• B. $\frac{3}{\pi }$
• C. $\frac{6}{\pi }$
• D. $6$

Answer 1

The correct option is B

$\frac{3}{\pi }$

Step 1. Given data

Motion of first particle is $y_1=0.25\sin \left(310t\right)$

Motion of second particle is $y_2=0.25\sin \left(316t\right)$

Step 2. Calculating frequency of two particles

By comparing the given equations with $y=A\sin \left(\omega t\right)$

For, first particle we get,

$$\begin{gathered} {\omega }_1=310 \\ 2\pi f_1=310 \\ {f}_1=\frac{310}{2\pi } \end{gathered}$$

Where, $f_1$ is the frequency of first particle

${\omega }_1$ is the angular frequency of first particle

Similarly for second particle, we get

$$\begin{gathered} {\omega }_2=316 \\ 2\pi f_2=316 \\ {f}_2=\frac{316}{2\pi } \end{gathered}$$

Where, $f_2$ is the frequency of second particle

${\omega }_2$ is the angular frequency of second particle

Step 3. Finding beat frequency.

We know, Beat frequency is given by

$$\begin{gathered} B=\left|f_2-f_1\right| \\ =\left|\left(\frac{316}{2\pi }\right)-\left(\frac{310}{2\pi }\right)\right| \\ =\left|\left(\frac{6}{2\pi }\right)\right| \\ =\frac{3}{\pi } \end{gathered}$$

Where, $f_1$ is the frequency of first particle

$f_2$ is the frequency of second particle

Hence, the correct option is (B).