Ch16-p062 - 62 Setting x = 0 in y = ym sin(k x − ω t φ gives y = ym sin(−ω t φ as the function being plotted in the graph We note that it

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Unformatted text preview: 62. Setting x = 0 in y = ym sin(k x − ω t + φ) gives y = ym sin(−ω t + φ) as the function being plotted in the graph. We note that it has a positive “slope” (referring to its tderivative) at t = 0: d y d ym sin(−ω t+ φ) = – ymω cos(−ω t+ φ) > 0 at t = 0. dt = dt This implies that – cos(φ) > 0 and consequently that φ is in either the second or third quadrant. The graph shows (at t = 0) y = 2.00 mm, and (at some later t) ym = 6.00 mm. Therefore, y = ym sin(−ω t + φ) |t = 0 φ = sin−1( 1 ) = 0.34 rad or 2.8 rad 3 (bear in mind that sin(θ) = sin(π − θ)), and we must choose φ = 2.8 rad because this is about 161° and is in second quadrant. Of course, this answer added to 2nπ is still a valid answer (where n is any integer), so that, for example, φ = 2.8 – 2π = −3.48 rad is also an acceptable result. ...
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This note was uploaded on 04/16/2011 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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