Question

The longitudinal waves travel in a coiled spring at a rate of 4m/s. The distance between two consecutive compressions is 20 cm. Find (1)- wavelength of longitudinal wave (2)- frequency of the wave motion.

Answer 1

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

Answer 2

(i) We can say the wavelength is around 0.2 m.

(ii) Therefore, wave frequency is 20 Hz

Explanation:

(i) The difference is equal to the wavelength between two successive compressions or unique factions.

Wavelength = 20 cm (given in the question)

We can say the wavelength is around 0.2 m.

(ii) Now, we have to determine the wave frequency.

STEP:1 Given values in the question

ν = 4 m/s

λ = 0.2 m

v is velocity and  λ is wavelength

STEP:2 We know that

f =V/ λ

put the given value in above equation

f = $\frac{4}{0.2}$ = 20 Hz

Therefore, wave frequency is 20 Hz

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