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The question is as follows:

A typical long-playing phonograph record (once known as an LP) plays for about $24$ minutes at $33 \frac{1}{3}$ revolutions per minute while a needle traces the long groove that spirals slowly in towards the center. The needle starts $5.7$ inches from the center and finishes $2.5$ inches from the center. Estimate the length of the groove.

Based on the given information, I was able to calculate the total number of revolutions -- which is $800$ revolutions ($24 \times \frac{100}{3}$). I don't know how to go further than that. Any help will be greatly appreciated.

geo_freak
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2 Answers2

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Hint: Instead of a spiral, suppose the groove on the LP were made of $800$ concentric circles of equal width; this will make calculations easier and will provide a very accurate approximation.

Since the groove has a non-zero width, the circles aren't really circles; they're annuli. Since you are given the outer and inner radii of the entire track, you can calculate the outer and inner radii of each annulus. Take the average of these to get the radius that the needle will trace, and use $C = 2\pi r$. Then find an efficient way to add up the $800$ lengths.

Here's the calculation for the outermost annulus. First, the width of each annulus will be $$w = \frac1{800}(5.7-2.5).$$ This means that the track inside that annulus will have a radius of $5.7 - \frac w2$, giving a length of $2\pi(5.7-\frac w2)$. Now only $799$ to go...

Théophile
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  • Thank you for the hint. I am still unsure about the concept of annuli and calculating the outer and inner radii. Can you please clarify that for me? – geo_freak Jun 02 '18 at 19:34
  • @geo_freak Sure, I'll show the calculation for the first one. – Théophile Jun 02 '18 at 19:38
  • @geo_freak In the calculation I added, note that I didn't quite follow my original hint: instead of taking the average of outer and inner radii, I started on the outside and went halfway in; this amounts to the same thing. – Théophile Jun 02 '18 at 19:47
  • So $\frac{w}{2}$ represents going halfway in, right? But why did you only decide to go halfway and not complete? – geo_freak Jun 02 '18 at 19:48
  • @geo_freak The outermost annulus has outer radius $5.7$ and inner radius $5.7-w$. Think of the annulus as a sort of racetrack, and the needle is a car that stays inside the track. The "middle of the road" is the average $\frac{5.7+(5.7-w)}2$, or $5.7-\frac w2$. – Théophile Jun 02 '18 at 19:58
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You can't assume that each revolution is exactly the same length! Each one will be slightly shorter than the one before. At best, you'd have to get the length of the outer-most groove and the inner most and figure out what the average of the two would be and then multiply by $800$, or whatever number you came up with for the amount of grooves per side. I was going for $20:00$ per side (most albums barely run $18:00$ per side, let alone $24:00$!) so, as the outer circumference is $37.700$ inches, I multiply that times $666$ because $33.3$ rpm $ \times 20:00$ is $666$. But, the results from these calculations are assuming that each revolution is exactly the same length, which they aren't! Dig! Get a full reel of recording tape ($7$" reel or $10$") and measure how much tape is used for just one revolution at the outer end of the tape. Now, get an empty reel and wrap one revolution of that same tape around the hub of the empty reel and measure that! It's nowhere NEAR as long as the first measurement you took! It's the same way with the grooves in a record.

Infinity_hunter
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Break-in Master
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