90.8k views
0 votes
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 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?

User Ragerory
by
6.1k points

1 Answer

3 votes

Answer:

197 hits/s

Step-by-step explanation:

If the record is turning at the rate of 33 rev/min that means it's turning


\omega = 33rev/min*2\pi rad/rev*(1)/(60)min/sec = 3.456 rad/s

If the groove is being played at radius of 10 cm (or 0.1m), that means the record is rotating with the speed of:


\omega*r = 3.456*0.1 = 0.346 m/s

The goove bumps is uniformly separated at 1.75 mm(or 0.00175m/bump) bumps. So the number of hits per second it should be


(0.346)/(0.00175) = 197.47 \approx 197 hits/s

User Artory
by
5.6k points