Answered

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

Answer :

okpalawalter8

Answer:

The rate at which the bump hit the style is  [tex]= 2516[/tex] 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 = [tex]14.6 * 100 = 146mm[/tex]

            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

                  [tex]C = 2 \pi r = 2 * 3.142 *147 = 924 \ mm[/tex]

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

                             [tex]n = \frac{C}{d} = \frac{924}{0.202} = 4574[/tex]

 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

                             [tex](33*4574)/60 = 2516[/tex]

   

Other Questions