Chapter 14 Homework 9 Solutions

# Chapter 14 Homework 9 Solutions - Solutions to Homework #9...

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Solutions to Homework #9 Chapter 14 Picture the Problem : The image shows two people talking on tin can telephone. The cans are connected by a 9.5- meter-long string weighing 32 grams. We wish to determine how the tension in the string affects the time for the message to travel across the string. Strategy : In problem 9, we found that the travel time across the string is given by / t md F = . Use this equation to calculate the time for the different tensions. Solution: 1. (a) Since the time is inversely related to the tension, increasing the tension will result in less time. 2. (b) Set the tension equal to 9.0 N: ( 29 ( 29 ( 29 0.032 kg 9.5 m / 9.0 N 0.18 s t = = 3. (c) Set the tension equal to 10.0 N: ( 29 ( 29 ( 29 0.032 kg 9.5 m / 10.0 N 0.17 s t = = Insight: As predicted, increasing the tension decreases the time for the message to travel the string. Picture the Problem : We are given the equation describing a wave and wish to determine the amplitude, wavelength, period, speed, and direction of travel. Strategy: The general form of a wave is given by 2 2 cos . y A x t T π λ = - ÷ Compare this equation to ( 29 15 cm cos , 5.0 cm 12 s y x t = - ÷ the equation given in the problem, to identify the parts of the wave. Use equation 14-1 and the definition of frequency to calculate the wave speed. Solution: 1. (a) Identify the amplitude as A: A = 15 cm 2. (b) Identify the wavelength as λ : 2 5.0 cm = , so 10.0 cm = 3. (c) Identify the period as T : 2 12 s T = , so T = 24 s 4. (d) Use equation14-1 to calculate the speed: 10 cm 0.42 cm/s 24 s v f T = = = = 5. (e) The wave travels to the right, because the t -term and x -term have opposite signs. Insight: The wave equation is a compact way of completely describing a wave, since it is possible to extract all of the wave properties from the equation. Picture the Problem : We need to calculate the wavelength of sound in air from its frequency.

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Strategy: Solve equation 14-1 for the wavelength, using 343 m/s for the speed of sound in air. Solution: 1. (a) Solve equation 14-1 for the wavelength: 343 m/s 0.807 m 425 Hz v f λ = = = 2. (b) Examine the relationship between wavelength and frequency: Wavelength is inversely related to frequency so, if the frequency increases the wavelength decreases. 3. (c) Calculate the wavelength at 450 Hz: 343 m/s 0.722 m 475 Hz = = Insight: As predicted, an increase in frequency corresponds to a decrease in wavelength. Chapter 15
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## Chapter 14 Homework 9 Solutions - Solutions to Homework #9...

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