1.

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 ?

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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