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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 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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MechEngineer

Answer:

197 hits/s

Explanation:

If the record is turning at the rate of 33 rev/min that means it's turning

[tex]\omega = 33rev/min*2\pi rad/rev*\frac{1}{60}min/sec = 3.456 rad/s[/tex]

If the groove is being played at radius of 10 cm (or 0.1m), that means the record is rotating with the speed of:

[tex]\omega*r = 3.456*0.1 = 0.346 m/s[/tex]

The goove bumps is uniformly separated at 1.75 mm(or 0.00175m/bump) bumps. So the number of hits per second it should be

[tex]\frac{0.346}{0.00175} = 197.47 \approx 197 hits/s[/tex]

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