A vinyl record is played by rotating the record so that an approximately circula

Question

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 rev/min, the groove being played is at a radius of 12.7 cm, and the bumps in the groove are uniformly separated by 0.184 mm. At what rate (hits per second) do the bumps hit the stylus?

Explanation / Answer

If the radius is 127 mm, then the circumference is equal to 127x2xp = 798 mm. Each bump is separated by 0.153 mm, so the number of bumps in the groove is equal to 798/0.153 = 5215 bumps. The record is spun at 33 rev/min, so the rate which the bumps hit the stylus is equal to (33x5215)/60 = 2870 hits per second. OR Distance travelled in 1 second = 2 . p . R . 33 / 60 = 43.89 cm Bumps run into in that distance = 438.9 mm / 0.153 mm = 2868 So the frequency heard = 2870 Hz

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