Thus we see that the block mass is inversely

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Thus, we see that the block mass is inversely proportional to the harmonic number squared. Thus, if the 447 gram block corresponds to harmonic number n then 447 286.1 = ( n + 1) 2 n 2 = n 2 + 2 n + 1 n 2 = 1 + 2 n + 1 n 2 . Therefore, 447 286.1 – 1 = 0.5624 must equal an odd integer (2 n + 1) divided by a squared integer ( n 2 ). That is, multiplying 0.5624 by a square (such as 1, 4, 9, 16, etc) should give us a number very close (within experimental uncertainty) to an odd number (1, 3, 5, …). Trying this out in succession (starting with multiplication by 1, then by 4, …), we find that multiplication by 16 gives a value very close to 9; we conclude n = 4 (so n 2 = 16 and 2 n + 1 = 9). Plugging m = 0.447 kg, n = 4, and the other values given in the problem, we find μ = 0.000845 kg/m = 0.845 g/m. 61. (a) The phasor diagram is shown here: y 1 , y 2 , and y 3 represent the original waves and y m represents the resultant wave. The horizontal component of the resultant is y mh = y 1 y 3 = y 1 y 1 /3 = 2 y 1 /3. The vertical component is y mv = y 2 = y 1 /2. The amplitude of the resultant is 2 2 2 2 1 1 1 1 2 5 0.83 . 3 2 6 m mh mv y y y y y y y = + = + = = (b) The phase constant for the resultant is 1 1 1 1 1 2 3 tan tan tan 0.644 rad 37 . 2 3 4 mv mh y y y y φ - - - = = = = = °
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CHAPTER 16 718 (c) The resultant wave is 1 5 sin ( 0.644 rad). 6 y y kx t ϖ = - + The graph below shows the wave at time t = 0. As time goes on it moves to the right with speed v = ϖ / k . 62. Setting x = 0 in y = y m sin ( k x - ϖ t + φ ) gives y = y m sin ( - ϖ t + φ ) as the function being plotted in the graph. We note that it has a positive “slope” (referring to its t - derivative) at t = 0: d y d t = d y m sin (-ϖ t+ φ 29 d t = – y m ϖ cos( t+ φ ) > 0 at t = 0 . 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 ) y m = 6.00 mm. Therefore, y = y m sin ( t + φ ) | t = 0 φ = sin - 1 ( 1 3 ) = 0.34 rad or 2.8 rad ( 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 2n π is still a valid answer (where n is any integer), so that, for example, φ = 2.8 – 2 π = - 3.48 rad is also an acceptable result. 63. We compare the resultant wave given with the standard expression (Eq. 16–52) to obtain ( 1 1 2 20m 2 / ,2 cos 3.0mm m k y - = = π λ = φ , and 1 2 0.820rad = φ . (a) Therefore, λ = 2 π / k = 0.31 m. (b) The phase difference is φ = 1.64 rad. (c) And the amplitude is y m = 2.2 mm.
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719 64. Setting x = 0 in a y = ϖ ² y (see the solution to part (b) of Sample Problem 16-2) where y = y m sin ( k x - ϖ t + φ ) gives a y = ϖ ² y m sin ( t + φ ) as the function being plotted in the graph. We note that it has a negative “slope” (referring to its t -derivative) at t = 0: d a y d t = d ( ϖ ² y m sin (-ϖ t+ φ 29 d t = y m ϖ 3 cos ( - ϖ t + φ ) < 0 at t = 0 . This implies that cos φ < 0 and consequently that φ is in either the second or third quadrant. The graph shows (at t = 0) a y = - 100 m/s², and (at another t ) a max = 400 m/s².
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