Problem_Set_#14

Anthony Putrino Problem Set # 14- 8, 14, 32, 36, 48, 66, 78 8. In shallow water of depth   d   the speed of waves is approximately   v=√ gd .   Find   (a)   the speed and   (b)   the period of a wave with a wavelength of 2.6 cm in water that is 0.75 cm deep. Solution: (a) Solve for wave speed. v = sqrt(gd) =sqrt((9.81m/s 2 )(.0075m) = .27m/s (b) Solve for wave period. v = λf = λ/T --> T = λ/v = .026m/.27m/s = .096sec 14. A brother and sister try to communicate with a string tied between two tin cans. If the string is 7.6 m long, has a mass of 29 g, and is pulled taut with a tension of 18 N, how much time does it take for a wave to travel from one end of the string to the other? Solution: Equate the time equal to distance divided by velocity to solve for time traveled. t = d/v = d(sqrt(u/F)) = sqrt(d 2 (m/d)F) = (sqrt(d 2 m/dF) = sqrt(dm/F) = sqrt((7.6m)(.029kg)/18N)) = .11sec 32. When you drop a rock into a well, you hear the splash 1.5 seconds later.   (a)   How deep is the well?   (b)   If the depth of the well were doubled, would the time required to hear the splash be greater than, less than, or equal to 3.0 seconds? Explain. Solution: (a) Solve for the falling time.

(1/2gt 1 2 = vt 2 )*2 = gt 1 2 = 2vt 2 d = .5g* t 1 2 d = 330* t 2 t 1 +t 2 =1.5 d = 300(1.5 - t 1 ) 300(1.5 - t 1 ) = .5g* t 1 2 --> 300(1.5 - t 1 ) = .5*9.8* t 1 2 t = 1.46sec d = .5*9.8* t 1 *2 = 10.44m or 11m (b) If the depth of the well were doubled, the time required to hear the splash would less than 3.0 seconds; this is because the fall time increases. 36. The distance to a point source is decreased by a factor of three.   (a)   By what multiplicative factor does the intensity increase?   (b)   By what additive amount does the intensity level increase? Solution: (a) Solve by calculating the ratio of intensities. I new /I old = (P/4πr new 2 )/(P/4πr old 2 ) = (r old /r new ) = (r old /1/3r old ) = 9 (b) Solve for intensity by substituting the new intensity level to find the change. β = 10log(I new /I 0 ) = 10log(9I old /I 0 ) --> 10log(I old /I 0 )+10log(9) = β old +10log(9) β new - β old = 10log(9) = +9.54dB