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प्रश्न
The longitudinal waves travel in a coiled spring at a rate of 4 m/s. The distance between two consecutive compressions is 20 cm. Find :
(i) Wavelength of the wave (ii) Frequency of the wave
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उत्तर
The distance between two consecutive compressions or rare factions is equal to its wavelength.
Hence, from the diagram of wave we can say that wavelength is 0.2 m.
(ii) Now we have to calculate the frequency of the wave.
Given: Velocity ν = 4 m/s
Wavelength λ = 0.2 m
We know the relation between velocity, frequency, and wavelength `f = v/lambda` ,
where v is the velocity of sound , f the frequency , and `lambda` the wavelength .
so ,
`f = 4/0.2` Hz
= 20 Hz
Therefore , frequency of the wave is 20 Hz .
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