MHT CET · Physics · Waves and Sound
Two waves \(Y_{1}=0 \cdot 25 \sin 316 \mathrm{t}\) and
\(\mathrm{Y}_{2}=0 \cdot 25 \sin 310 \mathrm{t}\)
are propagating along the same direction. The number of beats produced per second are
- A \(\frac{\pi}{2}\)
- B \(\frac{2}{\pi}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{3}{\pi}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{\pi}\)
Step-by-step Solution
Detailed explanation
No. of beats \(=\mathrm{n}_{1}-\mathrm{n}_{2}\)
\(\mathrm{Y}_{1}=0.25 \sin 316 \mathrm{t}\)
\(\therefore \frac{2 \pi}{\mathrm{T}}=316 \quad \therefore \mathrm{n}_{1}=\frac{316}{2 \pi}\)
\(\mathrm{Y}_{2}=0.25 \sin 310 \mathrm{t}\)
\(\therefore \mathrm{n}_{2}=\frac{310}{2 \pi}\)
\(\quad \mathrm{n}_{1}-\mathrm{n}_{2}=\frac{316-310}{2 \pi}=\frac{6}{2 \pi}=\frac{3}{\pi}\)
\(\mathrm{Y}_{1}=0.25 \sin 316 \mathrm{t}\)
\(\therefore \frac{2 \pi}{\mathrm{T}}=316 \quad \therefore \mathrm{n}_{1}=\frac{316}{2 \pi}\)
\(\mathrm{Y}_{2}=0.25 \sin 310 \mathrm{t}\)
\(\therefore \mathrm{n}_{2}=\frac{310}{2 \pi}\)
\(\quad \mathrm{n}_{1}-\mathrm{n}_{2}=\frac{316-310}{2 \pi}=\frac{6}{2 \pi}=\frac{3}{\pi}\)
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