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प्रश्न
A sound wave frequency 100 Hz is travelling in air. The speed of sound in air is 350 m s−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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उत्तर
Given:
Speed of sound in air v = 350 m/s
Frequency of sound wave f = 100 Hz
a) As we know,
\[v = f\lambda\]
\[\therefore \lambda = \frac{v}{f}\]
\[ \Rightarrow \lambda = \frac{350}{100} = 3 . 5 m\]
Distance travelled by the particle:
Δx = (350 × 2.5 × 10−3) m
Phase difference is given by:
\[\phi = \frac{2\pi}{\lambda} \times ∆ x\]
\[\text { On substituting the values we get: }\]
\[\phi = \left( \frac{2\pi \times 350 \times 2 . 5 \times {10}^{- 3}}{3 . 5} \right)\]
\[ \Rightarrow \phi = \left( \frac{\pi}{2} \right)\]
(b) For the second case:
Distance between the two points:
\[∆ x\]= 10 cm = 0.1 m
\[\Rightarrow \phi = \frac{2\pi}{\lambda} ∆ x\]
\[\text { On substituting the respective values in the above equation, we get: }\] \[\phi = \frac{2\pi \times 0 . 1}{3 . 5} = \frac{2\pi}{35}\]
The phase difference between the two points is \[\frac{2\pi}{35}\]
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