Question:
Two sources of sound are emitting progressive waves \(\mathrm{y}_1=4 \sin 710 \pi \mathrm{t}\) and \(\mathrm{y}_2=3 \sin 702 \pi \mathrm{t}\). The sources are placed close to each other. The number of beats heard per second and intensity ratio between waxing and waning are respectively
Two sources of sound are emitting progressive waves \(\mathrm{y}_1=4 \sin 710 \pi \mathrm{t}\) and \(\mathrm{y}_2=3 \sin 702 \pi \mathrm{t}\). The sources are placed close to each other. The number of beats heard per second and intensity ratio between waxing and waning are respectively
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Beat frequency is the difference of the two frequencies; the intensity ratio uses the sum and difference of the amplitudes squared.
Updated On: Nov 13, 2026
- \(4,16:9\)
- \(8,16:9\)
- \(4,49:1\)
- \(8,49:1\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
For \(y=A\sin 2\pi ft\), the angular frequency is \(\omega=2\pi f\). Beats per second equal \(f_1-f_2\).
Step 2: Beat frequency
\(f_1=\dfrac{710\pi}{2\pi}=355\text{ Hz}\) and \(f_2=\dfrac{702\pi}{2\pi}=351\text{ Hz}\), so beats per second \(=4\).
Step 3: Intensity ratio
Intensity is proportional to amplitude squared. Maximum amplitude is \(4+3=7\) and minimum is \(4-3=1\).
Step 4: Result
\(\dfrac{I_{max}}{I_{min}}=\dfrac{7^2}{1^2}=49:1\).
Final Answer:
There are 4 beats per second and the intensity ratio is 49:1. \[ \boxed{\text{(C) }4,\ 49:1} \]
For \(y=A\sin 2\pi ft\), the angular frequency is \(\omega=2\pi f\). Beats per second equal \(f_1-f_2\).
Step 2: Beat frequency
\(f_1=\dfrac{710\pi}{2\pi}=355\text{ Hz}\) and \(f_2=\dfrac{702\pi}{2\pi}=351\text{ Hz}\), so beats per second \(=4\).
Step 3: Intensity ratio
Intensity is proportional to amplitude squared. Maximum amplitude is \(4+3=7\) and minimum is \(4-3=1\).
Step 4: Result
\(\dfrac{I_{max}}{I_{min}}=\dfrac{7^2}{1^2}=49:1\).
Final Answer:
There are 4 beats per second and the intensity ratio is 49:1. \[ \boxed{\text{(C) }4,\ 49:1} \]
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