Question

Longitudinal waves travel at the speed of $10\mathrm{m}{\mathrm{s}}^{-1}$ in a coiled spring. If the distance between two successive rarefactions is $1.0\mathrm{cm}$, find the frequency of the wave.

Answer 1

Step 1: Given data

Speed($\mathrm{v}$) of longitudinal wave = $10\mathrm{m}{\mathrm{s}}^{-1}$

Distance between two successive rarefactions = $1.0\mathrm{cm}$

Step 2: Concept and Formula used:

1. Distance between two consecutive compressions or rarefaction is called wavelength($\lambda$).
2. $\mathrm{Frequency}\left(\mathrm{f}\right)\mathrm{of}\mathrm{the}\mathrm{wave}=\frac{\mathrm{Speed}\left(\mathrm{v}\right)}{\mathrm{Wavelength}\left(\mathrm{\lambda }\right)}$

Step 3: Calculation of wavelength and frequency:

Wavelength($\lambda$) = Distance between two consecutive rarefactions

Wavelength($\lambda$) = 1.0cm

Now,

$$\begin{gathered} \mathrm{f}=\frac{10\mathrm{m}{\mathrm{s}}^{-1}}{1\mathrm{cm}} \\ \mathrm{But},10\mathrm{m}{\mathrm{s}}^{-1}=10\times 100\mathrm{cm}{\mathrm{s}}^{-1} \\ =1000\mathrm{cm}{\mathrm{s}}^{-1} \\ \mathrm{f}=\frac{1000\mathrm{cm}{\mathrm{s}}^{-1}}{1\mathrm{cm}} \\ \mathrm{f}=1000\mathrm{Hz} \end{gathered}$$

Thus,

Frequency of the wave= $1000\mathrm{Hz}$