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

(i). The distance between two consecutive compressions or rarefactions is equal to its wavelength.

Hence, wavelength is$=20\ cm=0.20\ m$

(ii). Speed of wave $v=4\ m/s$

Wavelength $\lambda=0.20\ m$

Speed of wave$v=frequency(f)\times wavelength(\lambda)$

$4\ m/s=frequency(f)\times0.20\ m$

Or $frequency(f)=\frac{4}{0.20}$

Or $frequency(f)=20\ m/s$

Therefore, the frequency of the sound wave is $20\ m/s$.

Updated on: 10-Oct-2022

570 Views

Related Articles

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements