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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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  1. 28 February, 11:08
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    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
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