Question

The ripples in a certain groove 10.2 cm from the center of a 33$\frac{1}{3}$-rpm phonograph record have a wavelength of 1.55 mm. What will be the frequency of the sound emitted?

Short answer

The frequency of the sound emitted is $229.74\mathrm{H}\mathrm{z}$.

Step-by-step solution

Given data

The frequency of the phonograph record is $f=33\frac{1}{3}\mathrm{r}\mathrm{p}\mathrm{m}$.

The distance of the ripple from the center is $r=10.2\mathrm{c}\mathrm{m}$.

The wavelength of the ripple is $\lambda =1.55\mathrm{m}\mathrm{m}$.

Concepts

The relationship between frequency$\left(f\right)$, wavelength, and linear velocity$\left(v\right)$is given by,

$v=f\lambda$

Calculation

Converting the given frequency in rpm to angular velocity in rad/s, we get

$\begin{array}{rl}c\omega & =2\pi f\\ & =2\pi \times \left(33\frac{1}{3}\mathrm{r}\mathrm{p}\mathrm{m}\right)\times \left(\frac{1min}{60\mathrm{s}}\right)\\ & =2\pi \times \left(\frac{100}{3}\mathrm{r}\mathrm{p}\mathrm{m}\right)\times \left(\frac{1min}{60\mathrm{s}}\right)\\ & =3.4907\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\end{array}$

The linear velocity is given by,

$\begin{array}{rl}cv& =r\omega \\ & =10.2\times {10}^{-2}\mathrm{m}\times 3.4907\mathrm{r}\mathrm{a}\mathrm{d}/\mathrm{s}\\ & =0.3561\mathrm{m}/\mathrm{s}\end{array}$

Now, the frequency is given by,

$\begin{array}{rl}cf& =\frac{v}{\lambda }\\ & =\frac{0.3561\mathrm{m}/\mathrm{s}}{1.55\times {10}^{-3}\mathrm{m}}\\ & =229.74\mathrm{H}\mathrm{z}\end{array}$

Therefore, the frequency of the sound emitted is $229.74\mathrm{H}\mathrm{z}$.