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A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bumps in the groove run into the stylus, causing it to oscillate. The equipment converts those oscillations to electrical signals and then to sound. Suppose that a record turns at the rate of 33 1 3 rev/min, the groove being played is at a radius of 10.0 cm, and the bumps in the groove are uniformly separated by 1.75 mm. At what rate (hits per second) do the bumps hit the stylus?

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Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, [tex]\omega=33\frac{1}{3} rev/min[/tex] , radius of the record [tex]r=0.1m[/tex] and the distance between any two successive bumps on the groove as [tex]d=1.75mm[/tex].

The linear speed of the record in meters per second is:

[tex]v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\[/tex]

#From [tex]v[/tex] above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

[tex]Hits/second=v/d \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199[/tex]

Hence the bumps hit the stylus at around 199hit/s

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