ch16 - 1. (a) The motion from maximum displacement to zero...

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1. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period: 11 1.47Hz. 0.680s f T == = (c) A sinusoidal wave travels one wavelength in one period: 1.40m 2.06m s. 0.680s v T = λ
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2. (a) The angular wave number is 1 22 3.49m . 1.80m k ππ == = λ (b) The speed of the wave is () 1.80m 110rad s 31.5m s. vf ω λ = = =
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3. Let y 1 = 2.0 mm (corresponding to time t 1 ) and y 2 = –2.0 mm (corresponding to time t 2 ). Then we find kx + 600 t 1 + φ = sin 1 (2.0/6.0) and kx + 600 t 2 + = sin 1 (–2.0/6.0) . Subtracting equations gives 600( t 1 t 2 ) = sin 1 (2.0/6.0) – sin 1 (–2.0/6.0). Thus we find t 1 t 2 = 0.011 s (or 1.1 ms).
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4. Setting x = 0 in u = ω y m cos ( kx t + φ ) (see Eq. 16-21 or Eq. 16-28) gives u = y m cos ( t + ) 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 u d t = d (−ω y m cos (−ω t+ φ)) d t = y m ² sin ( t + ) > 0 at t = 0 . This implies that – sin > 0 and consequently that φ is in either the third or fourth quadrant. The graph shows (at t = 0) u = 4 m/s, and (at some later t ) u max = 5 m/s. We note that u max = y m . Therefore, u = u max cos ( t + ) | t = 0 ¡ = cos 1 ( 4 5 ) = ± 0.6435 rad (bear in mind that cos θ = cos( θ ) ), and we must choose = 0.64 rad (since this is about 37° and is in fourth quadrant). Of course, this answer added to 2n π is still a valid answer (where n is any integer), so that, for example, = 0.64 + 2 π = 5 .64 rad is also an acceptable result.
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5. Using v = f λ , we find the length of one cycle of the wave is λ = 350/500 = 0.700 m = 700 mm. From f = 1/ T , we find the time for one cycle of oscillation is T = 1/500 = 2.00 × 10 –3 s = 2.00 ms. (a) A cycle is equivalent to 2 π radians, so that /3 rad corresponds to one-sixth of a cycle. The corresponding length, therefore, is λ /6 = 700/6 = 117 mm. (b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2 π rad. Thus, the phase difference is (1/2)2 = rad.
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6. (a) The amplitude is y m = 6.0 cm. (b) We find λ from 2 π / λ = 0.020 : = 1.0×10 2 cm. (c) Solving 2 f = ω = 4.0 π , we obtain f = 2.0 Hz. (d) The wave speed is v = λ f = (100 cm) (2.0 Hz) = 2.0×10 2 cm/s. (e) The wave propagates in the – x direction, since the argument of the trig function is kx + t instead of kx t (as in Eq. 16-2). (f) The maximum transverse speed (found from the time derivative of y ) is () 1 max 24 . 0 s 6 .0cm 75cm s. m uf y = π = (g) y (3.5 cm, 0.26 s) = (6.0 cm) sin[0.020 π (3.5) + 4.0 π (0.26)] = –2.0 cm.
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7. (a) Recalling from Ch. 12 the simple harmonic motion relation u m = y m ω , we have 16 400rad/s. 0.040 == Since = 2 π f , we obtain f = 64 Hz. (b) Using v = f λ , we find λ = 80/64 = 1.26 m 1.3 m . (c) The amplitude of the transverse displacement is 2 4.0 cm 4.0 10 m. m y × (d) The wave number is k = 2 / = 5.0 rad/m. (e) The angular frequency, as obtained in part (a), is 2 16/ 0.040 4.0 10 rad/s. × (f) The function describing the wave can be written as () 0.040sin 5 400 yx t φ =− + where distances are in meters and time is in seconds. We adjust the phase constant to satisfy the condition y = 0.040 at x = t = 0. Therefore, sin = 1, for which the “simplest” root is = π /2. Consequently, the answer is 0.040sin 5 400 .
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ch16 - 1. (a) The motion from maximum displacement to zero...

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