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?

Question

Physics

Courtney Hill

28 February, 07:45

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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Mollie Donaldson · Answer 1

Given that,

Angular velocity w=33⅓ rev/min

Let convert to rad/sec

1 Revolution=2πrad

1 minute = 60 seconds

Then,

33⅓rev/min = 100/3 rev/mins * 2πrad/1 rev * 1min/60sec

Therefore, w=3.491rad/sec

Also, given that radius of grove is 10cm

Then, r=0.1m

The distance between each bump is 1.75mm

Also, d=1.75/1000=0.00175m

The linear velocity (v) of the record is given as

v=wr

v=3.491*0.1

v=0.3491 m/s

So, the number hit strike by the bump is given as

N = v/d

Since v=0.3491 and d=0.00175m

Then,

N=0.3491/0.00175

N=199.49hits/sec

The rate at which the bumps hit the stylus is 199.5hits/secs