problem set 14

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.  v is the speed of the wave,  g is the acceleration due to gravity (approximately 9.81 m/s^2),  d is the depth of the water = 0.0075m v = sqrt 9.81 m/s^2 x 0.0075m = 0.27 m/s So, the speed of the wave is approximately 0.271 m/s. The wavelength is given as 2.6 cm = 0.026m The formula would follow as f(frequency) = 0.271 m/s / 0.026m = 10.42 Hz So now to find the period of the wave it would follow as 1/f = 1/10.42 Hz = 0.096 s So, the period of the wave is approximately 0.096 s. 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? Given:  The value of mass given is 29 g.  The length of the string is 7.6 m.  The tension on the string is 18 N. Time = sqrt mass x length / tension in the string t = sqrt 29 x 10^-3 x 7.6 / 18 = sqrt 0.0122 = 0.11 s would be the amount of time to take for a wave to travel from one end to another.

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. Distance fallen by rock would be structured as: d = 0.5 gt^2  d is the depth of the well,  g is the acceleration due to gravity (approximately 9.81 m/s^2),  t is the time it takes for the sound to travel back up. t =1.5s d = 0.5 x 9.81 m/s^2 x (1.5s) ^2 = 11.025 m So, the depth of the well is approximately 11.025 m. And if the depth was doubled then it would be 22.05 m. To find time for the well being doubled, it would follow as: t = sqrt 2d/g t = 2 x 22.05m / 9.81 m/s^2 = 2.12 s So, if the depth of the well is doubled, the time required to hear the splash would be less than 3.0 s. 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? a) Using inverse square law: I ∝ 1/r 2. I= intensity and r = distance from source. If the distance is decreased by a factor of three: it would be = r = 1/3 I = 1 / 1/9r^2 = 9 So, the intensity increases by a factor of 9. b) To find the additive amount by which the intensity level increases in decibels, we use the logarithmic scale: Intensity   Level   (in   dB) =10 x log10 (I/I) Intensity   Level   (in   dB) =10 x 0.954 = 9.54 dB So, the intensity level increases by approximately 9.54 dB.

f 3 = 3 x 343 m/s / 2 x 0.44 m = 1029 / 0.88 = 1170.45 Hz. So, the frequency of the third harmonic of this bottle is approximately 1170.45 Hz. 78. To tune middle C on a piano, a tuner hits the key and at the same time sounds a 261-Hz tuning fork. If the tuner hears 3 beats per second, what are the possible frequencies of the piano key? The number of beats per second ( B ) is given by the difference in frequencies (f1 and f2). B = f1 – f2 - The tuner hits the piano key and sounds a 261-Hz tuning fork. - The tuner hears 3 beats per second. B= 261 – f2 = 3 To find f2: f 2 = 261 + 3 = 264Hz or f 2 = 261 – 3 = 258Hz Therefore, based on the given information and the occurrence of 3 beats per second, the possible frequencies of the piano key (f2) are 258 Hz and 264 Hz.