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Stan's Cassette (Posted on 2004-04-27) Difficulty: 3 of 5
Stan likes music, and likes oldies the best. In his collection of music he has a cassette tape of “The Cassettes 20 Greatest Hits”. The amazing thing about the 20 Greatest Hits, is that each hit is exactly 3 minutes long. This fits perfectly on a 60-minute tape.

Songs 1 – 10 are on Side A, while 11 – 20 are on Side B. When listening to the tape, Stan does one strange thing. He listens to the first 5 songs on side A, then flips it to the other side. He then listens to Side B from that point until the tape ends. How many songs does Stan hear?

  • The tape is wrapped around two spools that have identical diameters of 1 cm.
  • The spool that is receiving the tape rotates at a constant speed.
  • The tape, when located all on one spool, measures 5 cm in diameter.

See The Solution Submitted by Leming    
Rating: 3.0909 (11 votes)

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Solution solution | Comment 4 of 17 |

In a real cassette player, the tape moves at a constant speed past the read head, something like 1 3/4 inches per second.  However in this fictitious machine, the take-up reel moves at a constant angular speed.  As tape builds up on this spool, the speed of the tape past the read head increases, so that a song at the beginning of a given side takes up less space than a song at the end of the side.  When the tape is flipped over, after 15 minutes worth, less than half the tape has passed the read head, and so less than half the tape is available to play in that flipped state, even though half the songs, at the end of a side would require even more than half the tape.

The speed of the tape past the read head will be proportional to the effective diameter of the take-up spool.  That diameter multiplies five-fold from the beginning to end. The diameter is also proportional to the square root of the visible area of the spool (with tape).  Also, the effective area of the take-up spool increases linearly with the amount of tape taken up so far. So in arbitrary units of speed,

ds/dt = sqrt(A) = sqrt(1 + 24s)

where s is the amount of tape passed so far and t is the amount of time passed so far, on the current side.  Units are arbitrary, but we'll calibrate after the integration.

We get

dt/ds = (1+24s)^-(1/2)

t=2(1+24s)^(1/2) / 24 + c

t = (1 + 24s)^(1/2) / 12 - 1/12

where the constant of integration is such as to make t = 0 when s = 0.

Now we can calibrate this into units such that t = 1 and s = 1 after the whole side has played:

t = (1 + 24s)^(1/2) / 4  -  1/4

When t is 1/2, having played 5 songs or half the time duration of the side, s can be solved to be 1/3.  The first third of a side holds the first half of the time.  This leaves only the last third (length-wise) of the tape to play on the second side.  How long does this play?

That's a situation equivalent to 2/3 of the tape length (linear, not temporal) having passed by.  Substituting s = 2/3, t comes out to be (sqrt(17) - 1) / 4, or approximately .7807764, which leaves only .2192236 units of time still to play, or 6.5767 minutes, or 2.192 songs.  Added to the 5 songs already heard, Stan hears a total of 7.192 songs.

Edited on April 27, 2004, 8:38 pm
  Posted by Charlie on 2004-04-27 15:46:49

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