The phonograph record has a groove in which the needle would sit. As the record turns, a certain groove passes the needle at 25 cm/s. If the wiggles in the groove are 45 μm apart, what is the frequency of the sound that results?
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First of all, 25 cm/s is 0.25 m/s. Remember! Always convert into SI units before ANY calulation. Secondly, 45 micro metres is 45*10^-6m. So the grooves are 45*10^-6m apart.
So, forget the record, and imagine a straight line. Imagine a ball rolling along that straight line, and every time the ball rolls 45*10^-6m, I hit it. How many times would I hit it per second? Well, imagine I hit it every time it travelled 25 cm. In one second, i'd obviously hit it once. If it was travelling at 50 m/s, I'd hit it twice per second. This is given by "Number of hits=speed/distance", as number of hits when the ball is travelling at 50m/s (is 2) is given by 50/25=2.
So, number of hits per second (or the frequency) would be, again, the speed/distance per hit. This gives us 0.25/(45*10^-6) = 5555.5 (recurring), or 5555 and 5/9. The lowest number of sig figs the question gives us is 2, so leave as 5600 Hz.
P.S I had no idea in this question, I made some very extensive guess work.
So, forget the record, and imagine a straight line. Imagine a ball rolling along that straight line, and every time the ball rolls 45*10^-6m, I hit it. How many times would I hit it per second? Well, imagine I hit it every time it travelled 25 cm. In one second, i'd obviously hit it once. If it was travelling at 50 m/s, I'd hit it twice per second. This is given by "Number of hits=speed/distance", as number of hits when the ball is travelling at 50m/s (is 2) is given by 50/25=2.
So, number of hits per second (or the frequency) would be, again, the speed/distance per hit. This gives us 0.25/(45*10^-6) = 5555.5 (recurring), or 5555 and 5/9. The lowest number of sig figs the question gives us is 2, so leave as 5600 Hz.
P.S I had no idea in this question, I made some very extensive guess work.
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f = V/λ = 0.25/45E-6 = 5555 Hz