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?

The frequency of the phonograph record is \(f\; = \;33\frac{1}{3}\;{\rm{rpm}}\).

The distance of the ripple from the center is \(r\; = \;10.2\;{\rm{cm}}\).

The wavelength of the ripple is \(\lambda \; = \;1.55\;{\rm{mm}}\).

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

\(v = \;f\lambda \)

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

\(\begin{aligned}{c}\omega &= \;2\pi f\\ &= \;2\pi \times \;\left( {33\frac{1}{3}\;{\rm{rpm}}} \right) \times \;\left( {\frac{{1\;\min }}{{60\;{\rm{s}}}}} \right)\\ &= \;2\pi \times \;\left( {\frac{{100}}{3}\;{\rm{rpm}}} \right) \times \;\left( {\frac{{1\;\min }}{{60\;{\rm{s}}}}} \right)\\ &= \;3.4907\;{\rm{rad/s}}\end{aligned}\)

The linear velocity is given by,

\(\begin{aligned}{c}v &= \;r\omega \\ &= \;10.2 \times \;{10^{ - 2}}\;{\rm{m}} \times \;3.4907\;{\rm{rad/s}}\\ &= \;0.3561\;{\rm{m/s}}\end{aligned}\)

Now, the frequency is given by,

\(\begin{aligned}{c}f\; &= \;\frac{v}{\lambda }\\ &= \;\frac{{0.3561\;{\rm{m/s}}}}{{1.55\; \times \;{{10}^{ - 3}}\;{\rm{m}}}}\\ &= \;229.74\;{\rm{Hz}}\end{aligned}\)

Therefore, the frequency of the sound emitted is \(229.74\;{\rm{Hz}}\).