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16 .10.
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So we're given the equation for a transverse wave on a very long string.
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Y equals 6 sign of 0 .02 pi x plus 4 pi t.
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Units on these constants are omitted for brevity, but we're told that y and x are measured in centimeters and time t is in seconds.
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So we can figure out units for our answers.
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And there's a number of things that we're curious about, or at least that we're told, we need to find.
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Maybe we're not so curious.
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So the first one is the amplitude.
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Now this one is quite simple.
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The amplitude is of course just the maximum value that the displacement takes.
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We know that the sine function has a maximum value of plus or minus one.
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So the amplitude is just going to be what's out here in front of it.
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That's going to be six centimeters.
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Next we'd like the wavelength.
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So lambda the wavelength is 2 pi divided by the angular wave number.
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The angular wave number here is 0 .02 times pi.
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So just putting that in, we get that the wavelength is 100 centimeters.
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For the frequency, we know that the frequency is the angular frequency divided by 2 pi.
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The angular frequency here is 4 pi...