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A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is `350 ms^-1` (a) By how much is the phase changed at a given point in 2.5 ms ? (b) What is the phase difference at a given instant between two points separated by a distance of 10.0 cm along the direction of propagation ? |
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Answer» Correct Answer - A::B::C a. Here given `n=100, v=350 m/s` `rarr lamda(v/n)=350/100=3.5m` In 2.5 ms, the distance travelled by the particle is given by `Dx=(350xx2.5xx10^-3)` So, phase difference `phi=(2pi)/lamdaxxDx` `=((2pixx350xx2.5xx10^-3)/3.5)` `=(pi/2)` b. In the second case Given `/_eta=10 cm=10^-1m` `So, phi=(2pi)/x/_x` `=(2pixx10^-1)/(350/100)=(2pi)/35` |
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