What is the period of an old fashioned phonograph record that turns through 33 1/3 rotations per minute?

3 Responses to “What is the period of an old fashioned phonograph record that turns through 33 1/3 rotations per minute?”

1. 
   #1
   wjllope
   December 20th, 2009 3:47 am

   60 seconds/33.33 rotations = 1.80 seconds/rotation

   i.e. 33.33 rotations in one minute means 33.33 rotations in 60 seconds so one rotation takes [60/33.33] seconds

   the period is the time for one rotation (or cycle).

   cheers

   EDIT: the formula is in my post (several times)…
   References :

2. 
   #2
   (EMT)Barcham
   December 20th, 2009 4:35 am

   Period will be 1/rpms, so 1/33.33
   References :

3. 
   #3
   eyeonthescreen
   December 20th, 2009 5:03 am

   Period T = 1/f; where f = 33.33 rev per min; so T = 1/33.33 min per revolution. Change this to seconds, you know, 1 min = 60 sec. so 1/33.33 min = ? sec. You can do the math.

   But do you understand what a period is? A period is the length of time something takes to move through one cycle. Using your problem, the period is the length of time the record takes to go 1 full rotation. So if it goes 33.33 times around in a minute, then it would take 1/33.33 minutes to go around once…and that’s the period. And I did this without a formula.

   Learn the physics and you don’t need a formula to memorize.
   References :

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