Sound Waves Tutorial Questions

Sound Waves Tutorial Questions ± The Hearing of a Bat Description: ± Includes Math Remediation. Find the wavelength of a high-frequency sound wave emitted by a bat and the time needed to hear its echo. Bats are mainly active at night. They have several senses that they use to find their way about, locate prey, avoid obstacles, and "see" in the dark. Besides the usual sense of vision, bats are able to emit high-frequency sound waves and hear the echo that bounces back when these sound waves hit an object. This sonar-like system is called echolocation. Typical frequencies emitted by bats are between 20 and 200 kHz. Note that the human ear is sensitive only to frequencies as high as 20 kHz. A moth of length 1.0 is flying about 1.0 from a bat when the bat emits a sound wave at 80.0 . The temperature of air is about 10.0 . To sense the presence of the moth using echolocation, the bat must emit a sound with a wavelength equal to or less than the length of the insect. The speed of sound that propagates in an ideal gas is given by , where 273.15 is the ratio of heat capacities ( ), for air), is the absolute temperature in kelvins (which is equal to the Celsius temperature plus is the molar mass of the gas (for air, the average molar mass is ). Part A Find the wavelength of the 80.0- wave emitted by the bat. Express your answer in millimeters. Hint 1. Relating wavelength, frequency, and speed of a wave In periodic waves, the speed at which the wave pattern travels is given by , where is the wavelength and is the frequency of the wave. Hint 2. Find the speed of sound in air ), and is the universal gas constant ( Find the speed of sound in air at 10.0 . Express your answer in meters per second. ANSWER: = 338 ANSWER: = 4.23 Part B Will the bat be able to locate the moth despite the darkness of the night? ANSWER: yes no Part C How long after the bat emits the wave will it hear the echo from the moth? Express your answer in milliseconds to two significant figures. Hint 1. How to approach the problem After emitting the high-frequency sound, the bat waits for any echoes coming back from possible obstacles. Therefore the time needed to locate an obstacle depends on the speed of sound and the distance of the obstacle from the bat. Hint 2. Find the time needed for the sound wave to reach the moth How long does it take the sound wave to reach the moth? Express your answer in milliseconds to three significant figures. ANSWER: 2.96 The time elapsed from the emission of the sound to the detection of its echo is the time the sound wave takes to travel 1 and back. ANSWER: 5.9 Distance Costs Decibels Description: Find the change in decibals corresponding to the 1/r^2 decrease in intensity for a given increase in distance. The general formula for the number of decibels corresponding to a change in sound intensity to . Part A relative to the former intensity is Suppose that a sound has initial intensity of sound measured in decibels. This sound now increases in intensity by a factor . What is the new level ? Express your answer in terms of and the factor , using for the log base 10. Hint 1. Find the new decibel level The decibel is a logarithmic function, so multiplying the intensity by a factor follows the same rules as taking the log of a product. Express the log of in terms of the separate logs of and . ANSWER: = Hint 2. Find the decibel increase corresponding to a factor Find the change in decibels times corresponding to a multiplicative factor of change in intensity, by taking in the given formula to be . Express your answer, in decibels, in terms of , using for the logarithm of base 10. ANSWER: = ANSWER: = Part B Find the change in intensity of a point source of sound that results from moving the listener initially at a distance to a distance from the source . Express the change in intensity in terms of logarithm of and , using for the base 10. Hint 1. What is the dependence of intensity on distance? How does the intensity three dimensions)? ANSWER: of sound vary with the distance from the source to the listener as the sound spreads out from a source (in Hint 2. Find the ratio of intensities What is the ratio of the new to the old intensity? Give your answer in terms of and . ANSWER: = ANSWER: = Note that is negative if is greater than because the logarithm of a number less than 1 is negative. Part C If is 3.16 times as far away from the source as , by how many decibels will the sound at be changed relative to that at ? Give your answer numerically in decibels. ANSWER: = -10 A Pipe Filled with Helium Description: Given the fundamental frequency produced in a pipe filled with air, determine the fundamental frequency when air is replaced with helium at the same temperature. A certain organ pipe, open at both ends, produces a fundamental frequency of 280 in air. Part A If the pipe is filled with helium at the same temperature, what fundamental frequency and the molar mass of helium to be 4.00 will it produce? Take the molar mass of air to be 28.8 . Express your answer in hertz. Hint 1. How to approach the problem The fundamental frequency produced in a pipe, open at either both ends or only one end, depends on the speed of sound and the length of the pipe. Therefore, given the fundamental frequency produced by the pipe when it is filled with air at a certain temperature, you can find the length of the pipe. Once you determine the length of the pipe, then you can find the fundamental frequency produced by the same pipe when it is filled with a different gas. Note that, because helium is less dense than air and has a lower molar mass, sound waves propagate faster in helium than in air. Hint 2. Find the length of the pipe Find the length in air to be 344 of an open pipe that produces a fundamental frequency of 280 when it is filled with air. Take the speed of sound . Express your answer in meters. Hint 1. Fundamental frequency in pipes The fundamental frequency produced in a pipe depends on the length of the pipe and the speed of sound . A pipe that is open at both ends is called an open pipe; a pipe that is closed at one end is called a stopped (or closed) pipe. For an open pipe, the fundamental frequency is . ANSWER: = = 0.614 Hint 3. Find the speed of sound in helium Find the speed of sound in helium for the situation described in this problem. Don't forget that the helium is at the same temperature as the air. Take the speed of sound in air at this temperature to be 344 . The ratio of heat capacities for helium is . Express your answer in meters per second. Hint 1. Find the temperature of helium The air in the pipe is replaced with helium at the same temperature . What is the temperature of helium then? Take the speed of sound in air at this temperature to be 344 for the molar mass of air and and use 28.8 for the ratio of heat capacities for air. Express your answer in kelvins. ANSWER: = = 293 Hint 2. Speed of sound in an ideal gas The speed of sound in an ideal gas depends essentially on the temperature and the molar mass for a certain gas, the speed of sound in that gas is , where is the gas constant. ANSWER: = ANSWER: = 1010 of the gas. Given the ratio of heat capacities = = 821 Because helium is less dense than air and has a lower molar mass, sound waves propagate faster in helium than in air. Thus, the frequencies produced in the pipe when it is filled with helium are higher than those produced in the same pipe filled with air. Part B Now consider a pipe that is stopped (i.e., closed at one end) but still has a fundamental frequency of 280 Part A, in air. How does your answer to , change? Hint 1. How to approach the question In general, the fundamental frequency produced in a pipe does depend on whether the pipe is open or stopped. For a stopped pipe, the fundamental frequency is given by . But, consider the ratio of the fundamental frequency in air to the fundamental frequency in helium: . Does this ratio depend on whether the pipe is open or stopped? ANSWER: increases. decreases. stays the same. The fundamental frequency of the pipe in helium is given by . This relationship is independent of the length of the pipe or whether the pipe is open or stopped. A relationship of this type, known as a scaling law, is very powerful because it allows you to solve problems without knowing all of the values that would normally be relevant. In this case, you can determine the frequency in helium knowing only the frequency in air and the ratio of the speed of sound in helium to that in air. Two Traveling Waves Beating Together Description: Show that superposition of two travelling waves can create beats that travel and find the group velocity. Learning Goal: To see how two traveling waves of nearly the same frequency can create beats and to interpret the superposition as a "walking" wave. Consider two similar traveling transverse waves, which might be traveling along a string for example: and They are similar because we assume that and are nearly equal and also that and . are nearly equal. Part A Which one of the following statements about these waves is correct? ANSWER: Both waves are traveling in the direction. Both waves are traveling in the direction. Only wave is traveling in the direction. Only wave is traveling in the direction. The principle of superposition states that if two waves each separately satisfy the wave equation then the sum (or difference) also satisfies the wave equation. This follows from the fact that every term in the wave equation is linear in the amplitude of the wave. Consider the sum of the two waves given in the introduction, that is, . These waves have been chosen so that their sum can be written as follows: where is a constant, and the functions and are trigonometric functions of first function, called the envelope, is a slowly varying function of both position ( and . This form is especially significant because the ) and time ( ), whereas the second varies rapidly with both position ( ) and time ( ). Traditionally, the overall amplitude is represented by the constant and , while the functions are trigonometric functions with unit amplitude. Part B Find , , and . Express your answer in terms of , term) varies slowly whereas amplitude. (the third term) varies quickly. Both , , , , , and . Separate the three terms with commas. Recall that and (the second should be trigonometric functions of unit Hint 1. A useful trigonometric identity Hint 2. Which is the envelope and which is the carrier wave? Recall that the carrier signal varies much more quickly over time and space than does the envelope signal. Given a choice between and , where , which wave oscillates faster in space? ANSWER: ANSWER: , , = , , Part C Which of the following statements about is correct? ANSWER: It is a rapidly oscillating wave traveling in the direction. It is a rapidly oscillating wave traveling in the direction. It is slowly oscillating in time but is standing still. It is traveling rapidly but oscillating slowly. Part D Which of the following statements about is correct if you assume that and are both positive? ANSWER: It is a slowly oscillating wave traveling in the direction. It is a slowly oscillating wave traveling in the direction. It is slowly oscillating in time but is standing still. It is traveling rapidly but oscillating slowly. Part E The envelope function can be written simply in terms of propagation of the envelope? Express your result in terms of and . and . If you do so, what is , the velocity of Hint 1. Substituting Substitute and and Express in terms of into the expression for and . . ANSWER: = Hint 2. Traveling wave velocity What is the velocity of a traveling wave Express your answer in terms of and ? . ANSWER: = ANSWER: = is called the group velocity because this ratio is the velocity at which the group of waves under a maximum of the envelope function appears to travel. It is technically defined as a derivative (although you found it for a finite difference): . This contrasts with the velocity of propagation of the waves themselves: . Information (e.g., applied as amplitude modulation of the wave, that is, as a variation in the envelope) travels at the group velocity. No information can be sent at the phase velocity of a wave, which therefore can exceed the speed of light. (Einstein's Special Theory of Relativity implies that neither a physical body nor information can travel faster than light.) The Doppler Effect on a Train Description: Find the frequency heard by a listener in motion with respect to the source of sound that is itself moving. A train is traveling at 30.0 relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 The speed of sound in air should be taken as 344 . . Part A What frequency is heard by a passenger on a train moving at a speed of 18.0 relative to the ground in a direction opposite to the first train and approaching it? Express your answer in hertz. Hint 1. How to approach the problem The listener is in motion with respect to the source of sound; therefore, you need to consider the Doppler effect. Moreover, you also need to take into account that the source of sound is moving. If you have a formula that gives the frequency shift when both source and listener are moving, you can apply it directly. If you only have formulas for the Doppler shift when either the source or listener is moving but not both, you will need to calculate the frequency shift in two steps. First, consider a point somewhere between the two trains and compute the frequency of the train whistle that would be heard by a stationary listener at that point. Then consider that point to be the (stationary) source of the sound that is observed by the passenger moving in the second train. Hint 2. Doppler shift equations for moving source or observer When a source of sound with frequency frequency moves at speed with respect to a stationary observer, the observer will hear a sound of given by . When a source of sound with frequency sound of frequency is stationary with respect to an observer moving with speed , the observer will hear a given by . In both cases, is the speed of sound and the sign to use depends on whether the source is moving toward or away from the observer. Hint 3. Doppler equations when both the source and the listener are in motion If is the frequency of a sound wave emitted by a source in motion at speed listener in motion at speed , the frequency of the sound wave heard by a is given by , where is the speed of sound. Hint 4. Determine the appropriate signs In the Doppler equation for a moving source and a moving observer, , where is the speed of sound, which signs belong in the numerator and denominator if the source and the observer are both moving toward each other? Recall that speed and frequency are both scalar quantities. Hint 1. Qualitative understanding of the Doppler shift If the source and listener are approaching each other, the frequency observed will be higher than the frequency emitted. Similarly, if the the source and listener are receding from each other, the frequency observed will be lower than the frequency emitted. ANSWER: A minus sign in the numerator and a minus sign in the denominator. A plus sign in the numerator and a plus sign in the denominator. A plus sign in the numerator and a minus sign in the denominator. A minus sign in the numerator and a plus sign in the denominator. Note that these are the signs that should be used when solving for . ANSWER: = 302 Part B What frequency is heard by a passenger on a train moving at a speed of 18.0 first train and receding from it? Express your answer in hertz. relative to the ground in a direction opposite to the Hint 1. How to approach the problem The listener is in motion with respect to the source of sound; therefore, you need to consider the Doppler effect. Moreover, you also need to take into account that the source of sound is moving. If you have a formula that gives the frequency shift when both source and listener are moving, you can apply it directly. If you only have formulas for the Doppler shift when either the source or listener is moving but not both, you will need to calculate the frequency shift in two steps. First, consider a point somewhere between the two trains and compute the frequency of the train whistle that would be heard by a stationary listener at that point. Then consider that point to be the (stationary) source of the sound that is observed by the passenger moving in the second train. Hint 2. Doppler shift equations for moving source or observer When a source of sound with frequency frequency moves at speed with respect to a stationary observer, the observer will hear a sound of , given by . When a source of sound with frequency sound of frequency is stationary with respect to an observer moving with speed , the observer will hear a given by . In both cases, observer. is the speed of sound and the sign to use depends on whether the source is moving toward or away from the Hint 3. Doppler equations when both the source and the listener are in motion If is the frequency of a sound wave emitted by a source in motion at speed listener in motion at speed , the frequency of the sound wave heard by a is given by , where is the speed of sound. Hint 4. Determine the appropriate signs In the Doppler equation for a moving source and a moving observer, , where is the speed of sound, which signs belong in the numerator and denominator if the source and the observer are both moving away from each other? Recall that speed and frequency are both scalar quantities. Hint 1. Qualitative understanding of the Doppler shift If the source and listener are approaching each other, the frequency observed will be higher than the frequency emitted. Similarly, if the the source and listener are receding from each other, the frequency observed will be lower than the frequency emitted. ANSWER: A minus sign in the numerator and a minus sign in the denominator. A plus sign in the numerator and a plus sign in the denominator. A plus sign in the numerator and a minus sign in the denominator. A minus sign in the numerator and a plus sign in the denominator. Note that these are the signs that should be used when solving for ANSWER: . = 228