Question

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

A
Zero
B
π
C
π/2
D
π/4

Solution

## The correct option is D $$\pi / 2$$Time interval between two consecutive beats$$T = \dfrac{1}{n_1 - n_2} = \dfrac{1}{260 - 256} = \dfrac{1}{4}$$sec so, $$t = \dfrac{1}{16} = \dfrac{T}{4}$$ secBy using time difference $$= \dfrac{T}{2 \pi} \times$$ Phase difference$$\Rightarrow \dfrac{T}{4} = \dfrac{T}{2 \pi} \times \phi$$$$\Rightarrow \phi = \dfrac{\pi}{2}$$Physics

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