{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Chapter-16 - Chapter 16 1(a The angular wave number is k =...

This preview shows pages 1–4. Sign up to view the full content.

693 Chapter 16 1. (a) The angular wave number is 1 2 2 3.49m . 1.80m k - π π = = = λ (b) The speed of the wave is ( ( 1.80m 110rad s 31.5m s. 2 2 v f ϖ λ = λ = = = π π 2. The distance d between the beetle and the scorpion is related to the transverse speed t v and longitudinal speed v l as t t d v t v t = = l l where t t and t l are the arrival times of the wave in the transverse and longitudinal directions, respectively. With 50 m/s t v = and 150 m/s v = l , we have 150 m/s 3.0 50 m/s t t t v t v = = = l l . Thus, if 3 3 3.0 2.0 4.0 10 s 2.0 10 s , t t t t t t t t - - ∆ = - = - = = × = × l l l l l then 3 (150 m/s)(2.0 10 s) 0.30 m 30 cm. d v t - = = × = = l l 3. (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: 1 1 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 4. (a) The speed of the wave is the distance divided by the required time. Thus,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CHAPTER 16 694 853 seats 21.87 seats/s 22 seats/s 39 s v = = . (b) The width w is equal to the distance the wave has moved during the average time required by a spectator to stand and then sit. Thus, (21.87 seats/s)(1.8 s) 39 seats w vt = = . 5. 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 + φ = 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). 6. Setting x = 0 in u = - ϖ y m cos ( k x - ϖ t + φ ) (see Eq. 16-21 or Eq. 16-28) gives u = - ϖ y m cos ( - 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 u d t = d (-ϖ y m cos (-ϖ t+ φ 29 d t = - y m ϖ ² sin ( - ϖ t + φ ) > 0 at t = 0 . This implies that – sin φ > 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 ϖ . Therefore, u = - u max cos ( - ϖ t + φ ) | t = 0 φ = cos - 1 ( 4 5 29 = ± 0.6435 rad ( bear in mind that cos θ = cos( - θ ) ) , and we must choose φ = - 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, φ = - 0.64 + 2 π = 5 .64 rad is also an acceptable result.
695 7. 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 π /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 π = π rad. 8. (a) The amplitude is y m = 6.0 cm. (b) We find λ from 2 π / λ = 0.020 π : λ = 1.0×10 2 cm. (c) Solving 2 π f = ϖ = 4.0 π , we obtain f = 2.0 Hz.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}