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Standard XII

Physics

Beat Frequency

The frequenci...

Question

The frequencies of two sound sources are $256$ $H z$ and $260 H z$ . At $t = 0,$ the intensity of sound is maximum. Then the phase difference at the time $t = 1 / 16$ sec will be

Zero

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π

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π / 2

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π / 4

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Solution

The correct option is D $π / 2$
Time interval between two consecutive beats
$T = \frac{1}{n_{1} - n_{2}} = \frac{1}{260 - 256} = \frac{1}{4}$ sec

so, $t = \frac{1}{16} = \frac{T}{4}$ sec

By using time difference $= \frac{T}{2 π} \times$ Phase difference

$\Rightarrow \frac{T}{4} = \frac{T}{2 π} \times ϕ$

$\Rightarrow ϕ = \frac{π}{2}$

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The frequencies of two sound sources are 256 Hz and 260Hz . At t=0, the intensity of sound is maximum. Then the phase difference at the time t=1/16 sec will be

The correct option is D π/2Time interval between two consecutive beatsT=1n1−n2=1260−256=14sec so, t=116=T4 secBy using time difference =T2π× Phase difference⇒T4=T2π×ϕ⇒ϕ=π2

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