MHT CET · Physics · Waves and Sound
Two waves \(Y_1=0.25 \sin 316 t\) and \(\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}\) are propagating along the same direction. The number of beats produced per second are
- A \(\frac{\pi}{3}\)
- B \(\frac{3}{\pi}\)
- C \(\frac{2}{\pi}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{\pi}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& y=A \sin \omega t \\
& \text {Given, } y_1=0.25 \sin 316 \mathrm{t} \\
& \text { Comparing with (i), } \omega_1=316 \\
& \omega_1=2 \pi \mathrm{f}_1 \\
\therefore \quad & \frac{316}{2 \pi}=\mathrm{f}_1 \\
& \text { Given, } \mathrm{y}_2=0.25 \sin 310 \mathrm{t} \\
& \text { Comparing with (i), } \omega_2=310 \\
& \omega_2=2 \pi \mathrm{f}_2 \\
\therefore \quad & \frac{310}{2 \pi}=\mathrm{f}_2
\end{array}\)
Number of beats produced per second is
\(=f_1-f_2=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{3}{\pi}\)
& y=A \sin \omega t \\
& \text {Given, } y_1=0.25 \sin 316 \mathrm{t} \\
& \text { Comparing with (i), } \omega_1=316 \\
& \omega_1=2 \pi \mathrm{f}_1 \\
\therefore \quad & \frac{316}{2 \pi}=\mathrm{f}_1 \\
& \text { Given, } \mathrm{y}_2=0.25 \sin 310 \mathrm{t} \\
& \text { Comparing with (i), } \omega_2=310 \\
& \omega_2=2 \pi \mathrm{f}_2 \\
\therefore \quad & \frac{310}{2 \pi}=\mathrm{f}_2
\end{array}\)
Number of beats produced per second is
\(=f_1-f_2=\frac{316}{2 \pi}-\frac{310}{2 \pi}=\frac{3}{\pi}\)
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