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 11.5 cm, and the bumps in the groove are uniformly separated by 0.374 mm. At what rate (hits per second) do the bumps hit the stylus?

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opudodennis opudodennis · Answer 1 · 2020-04-03T05:06:07+02:00

Answer:

1063 hits/sec

Explanation:

-Angular velocity of the record is given as:

[tex]\omega= 33 \ rev/min=0.55 \ rev/s[/tex]

-Radius of the record is given as:

[tex]r=11.5\ cm=0.115\ m[/tex]

The linear speed of the record is calculated as:

[tex]v=\omega r\\\\=0.55\times 2\pi \times 0.115\\\\=0.3974 \ m/s[/tex]

The grove bumps are uniformly separated as 0.374 mm(0.000374), the number of hits per second is calculated as:

[tex]hits/s=\frac{v}{us}\\\\=\frac{0.3974}{0.000374}\\\\=1062\ hits/s\approx 1063 \ hits/s[/tex]

Hence, there are 1063 hits/sec