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A golden oldie


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Johnny was going through his collection of CD's when grandpa came over with his favorite 33 1/3 rpm long-play phonograph record. Johnny was fascinated with the ancient technology - imagine a needle playing the sound in all those grooves by actually touching them the record!

Hoping Johnny could be impressed with the ancient method of storing music, grandpa brought with him an equally old microscope and challenged Johnny to count the grooves. Assuming there was precisely 50 minutes of music on the record, how many grooves did Johnny count?

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Assuming that the record was a standard 12" with a 4" middle section then the math is quite easy, there is 8" of diameter to play 50 minutes of music, so the answer is obviously.....

I bet you thought you got me.

He should have counted 2 grooves one on each side.

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Okay, now that the intended solution has been given:

What if Johnny is counting revolutions of the record and not grooves, and he only listens to 25 minutes of music? How many revolutions did he count? Does it matter if he listened to the first 25 minutes or the last half of the record?

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What if Johnny is counting revolutions of the record and not grooves, and he only listens to 25 minutes of music? How many revolutions did he count?
25x33 1/3 = 833 1/3 revolutions.
Does it matter if he listened to the first 25 minutes or the last half of the record?

[1] No, becasue there were exactly 25 minutes of music on each side.

[2] No, because wherever the music is, the record spins at the same speed.

The vibrations are more closely spaced near the center of the record.

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Assuming that the record was a standard 12" with a 4" middle section then the math is quite easy, there is 8" of diameter to play 50 minutes of music, so the answer is obviously.....

I bet you thought you got me.

He should have counted 2 grooves one on each side.

OK, so you counted correctly; but let's ask the companion question: given everything that's been said, and simplifying things by assuming now that all 50 minutes of music are on side 1 of the record, and further simplifying by assuming equal spacing of the grooves, approximating pi by 3.14 or 22/7, whichever is easier to use, and rounding your answer to the nearest inch ...

what is the total distance traveled by the needle, in 50 minutes, from the first note of the music to the last?

[edited, to point out the question ...]

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Isn't the answer 1? There would only be 1 groove no matter how many minutes were played. If you were asking how many ridges would encounter moving from the outside to the inside of the record, it would be 50x33.3333333 = 1500 + 150 + 15 + 1.5 + 0.15... = 1666.6667 ridges. (1/2 this is you said it was really 25 minutes per side).

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I don't like this question and I can't see a clearly defined answer in this forum... AM I missing a post that says ANSWER= _____, or is the highlighting in the first post supposed to indicate that the answer is in the question or something?

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I don't like this question and I can't see a clearly defined answer in this forum... AM I missing a post that says ANSWER= _____, or is the highlighting in the first post supposed to indicate that the answer is in the question or something?

There are 2 grooves on the record, one on each side.

When a record plays, the needle continues down the same groove, otherwise you'd have to lift the needle ~833 times for each side of the record. :P

Edited by miya
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1 groove on each side. Looking at a record would prove it, because, as Miya said, you would need to lift the needle every time it made a revolution in order to play a full side if there were multiple grooves.

A little note, at the end of a record (middle), there is a never ending circle, so in reality there is no real end point for the needle.

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the phonograph spins at 33.333 revolutions in one minute, which means 33.333 grooves get traversed in 1 minute.

if there are 50 minutes of play on the record, this translates to their being 50 * 33.3333 = 1666.66665 grooves on the record.

for this to be correct, the record should also play 1 minute worth of sound at 33 1/3 revolutions.

I am not sure what the 0.6666 groove represents though.

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the phonograph spins at 33.333 revolutions in one minute, which means 33.333 grooves get traversed in 1 minute.

if there are 50 minutes of play on the record, this translates to their being 50 * 33.3333 = 1666.66665 grooves on the record.

for this to be correct, the record should also play 1 minute worth of sound at 33 1/3 revolutions.

I am not sure what the 0.6666 groove represents though.

He he!! - great math - how does the needle go from one grove to the other ?? does it have the same on the other side ?? :o:o:o = O+M+G !!!
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He he!! - great math - how does the needle go from one grove to the other ?? does it have the same on the other side ?? :o:o:o = O+M+G !!!

there was no mention of the record having grooves on both sides. but if that is case, and if we are to count the grooves as one on each side since the groove is connected from the inside all the way to the outside (as opposed to counting the grooves in a straight line) then there are two grooves on the record.

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I don't like this question and I can't see a clearly defined answer in this forum... AM I missing a post that says ANSWER= _____, or is the highlighting in the first post supposed to indicate that the answer is in the question or something?

I understand that the expected answer is one on each side, but there have been 33 1/3 LPs with one blank side and even some with more than one groove. Yes, more than one groove. When playing such a record, you never knew which groove you will get.

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