MHT CET · Physics · Waves and Sound
Two waves \(Y_1=0.25 \sin 316 \mathrm{t}\) and \(\mathrm{Y}_2=0.25 \sin 310 \mathrm{t}\) are propagating in same direction. The number of beats produced per second are
- A \(\frac{3}{\pi}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{2}{\pi}\)
Answer & Solution
Correct Answer
(A) \(\frac{3}{\pi}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{y}_1=0.25 \sin 316 \mathrm{t}, \mathrm{y}_2=0.25 \sin 310 \mathrm{t} \\ & \therefore 2 \pi \mathrm{f}_1=316 \text { and } 2 \pi \mathrm{f}_2=310 \\ & \therefore \mathrm{f}_1=\frac{316}{2 \pi}=\frac{158}{\pi} \\ & \mathrm{f}_2=\frac{310}{2 \pi}=\frac{155}{\pi} \\ & \therefore \text { Beat frequency }=\mathrm{f}_1-\mathrm{f}_2 \\ & =\frac{158}{\pi}-\frac{155}{\pi}=\frac{3}{\pi}\end{aligned}\)
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