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 gr

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 14.6 cm, and the bumps in the groove are uniformly separated by 0.202 mm. At what rate (hits per second) do the bumps hit the stylus

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Thông Đạt 5 years 2021-08-24T22:22:22+00:00 1 Answers 141 views 1

Answers ( )

  1. lethu
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    2021-08-24T22:23:45+00:00
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    Answer:

    The rate at which the bump hit the style is  = 2516 hits per second

    Explanation:

    From the question we are given that

                 Velocity of the record is 33 rev/min

                 The radius of the grove  14.6 cm = 14.6 * 100 = 146mm

                distance of  separation  of the bumps in the groove  is d =  0.202 mm

    Now since we know the radius to obtain the circumference of the record would be

                      C = 2 \pi r = 2 * 3.142 *147 = 924 \ mm

     Since each of the bump is separated from one another by 0.202 mm the number of bumps can be obtained mathematically as

                                 n = \frac{C}{d} = \frac{924}{0.202} = 4574

     Where n is the number of bumps

                 C is the circumference of the record

                  d is the distance between bumps

    We are told from the question that the rate a record turn is 33rev/min

    Hence the rate at which the bumps hit the style would be

            =  Rate at which record turn in seconds × The number of bumps

    Since 1 minute is equal to 60 seconds

      This means each and every bump  would hit the style  at the rate of

                                 (33*4574)/60 = 2516

       

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