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Question

Two sources of sound A and B produce waves of 350 Hz. The particle at point P is vibrating under the influence of these two waves. The amplitudes of the two waves are 0.3 mm and 0.4 mm. If APBP=25 cm and velocity of sound is 350 m/s, the resultant amplitude of the particle at point P will be

A
0.7 mm
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B
0.1 mm
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C
0.2 mm
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D
0.5 mm
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Solution

The correct option is D 0.5 mm
We know that ,
Wavelength, λ=vf=350350=1 m or 100 cm

Also, path difference (Δx) between the waves at the point of observation is APBP=25 cm.
Hence, phase difference Δϕ=2πλ×(Δx)=2π100×25=π2

Amplitude of resultant wave
A=a21+a22
From the data given in the question,
A=(0.3)2+(0.4)2=0.5 mm

Hence, option (d) is the correct answer.

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