Question:
Two waves \(Y_1 = 0.25sin316t\) and \(Y_2 = 0.25sin310t\) are propagating along the same direction. The number of beats produced per second are
Two waves \(Y_1 = 0.25sin316t\) and \(Y_2 = 0.25sin310t\) are propagating along the same direction. The number of beats produced per second are
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Beat frequency is the difference of the frequencies, and each frequency is omega over 2 pi.
Updated On: Oct 1, 2026
- \(\frac{3}{π}\)
- \(\frac{π}{3}\)
- \(\frac{2}{π}\)
- \(\frac{π}{2}\)
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The Correct Option is A
Solution and Explanation
Step 1: Frequencies
\(\omega = 2\pi f\), so \(f_1 = \frac{316}{2\pi}\) and \(f_2 = \frac{310}{2\pi}\).
Step 2: Beat frequency
\(f_1 - f_2 = \frac{316-310}{2\pi} = \frac{6}{2\pi} = \frac3\pi\). Option (A).
Final Answer:
The number of beats per second is 3/pi. \[ \boxed{\text{(A)}\ \frac3\pi} \]
\(\omega = 2\pi f\), so \(f_1 = \frac{316}{2\pi}\) and \(f_2 = \frac{310}{2\pi}\).
Step 2: Beat frequency
\(f_1 - f_2 = \frac{316-310}{2\pi} = \frac{6}{2\pi} = \frac3\pi\). Option (A).
Final Answer:
The number of beats per second is 3/pi. \[ \boxed{\text{(A)}\ \frac3\pi} \]
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