Question

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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 :
(a) Wavelength of the wave
(b) Frequency of the wave
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Answer 1

$\underline{\fcolorbox{red}{pink}{ \huge{Solution :}}}$

a) The distance between two consecutive compressions is equal to it's wavelength

Thus , the wavelength of the wave is 0.2 m { 1 cm = 1/100 m }

b) We know that , the relationship between frequency and wavelength is given by

$\large \mathtt{\fbox{frequency = \frac{speed \: of \: wave}{wavelength} }}$

Substitute the known values , we get

$\sf \mapsto frequency = \frac{4}{0.2} \\ \\\sf \mapsto frequency = \frac{40}{2} \\ \\ \sf \mapsto frequency = 20 \: \: hz$

Hence , the frequency of the wave is 20 hz

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Answer 2

Explanation:

Given data :

As we know that from given data,

v = 4m/s

Distance among two consecutive is obviously the wavelength of longitudinal wave

λ = 10 cm = 10 / 100 = 0.1 m

Solution :

Frequency of wave - motion = v / λ = 4 / 0.10  

Frequency = 40 Hz