Question 01. A. Incorrect! The speed of sound is not the same in all medium; it is dependent on the properties of the material.

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1 High School Physics - Problem Drill 15: Sound 1. Which of these is not a true statement about sound waves? Question 01 (A) Sound waves are travel at different speeds in different mediums. (B) Sound waves can be heard in space. (C) Sound waves carry energy. (D) Sound waves can be heard under water. (E) Sound waves do not transport matter from one place to another. The speed of sound is not the same in all medium; it is dependent on the properties of the material. B. Correct! Sound waves require a medium through which to travel. In space there may be a few atoms and molecules but they are so widely spaced a wave cannot be detected and space is essentially a vacuum through which no sound travels. Sound is a mechanical wave, consider the fact that molecules of medium are being moved, to do this requires energy, as the wave progresses so do the energy. Sound waves can travel through the medium of water. If you have ever put your head under water you will know that you can still hear. The molecules of the medium are being displaced about an equilibrium position. Once the wave has passed the molecules will return to this position. The speed of sound is not the same in all medium; it is dependent on the properties of the material. Sound waves require a medium through which to travel. In space there may be a few atoms and molecules but they are so widely spaced a wave cannot be detected and space is essentially a vacuum through which no sound travels. Sound is a mechanical wave, consider the fact that molecules of medium are being moved, to do this requires energy, as the wave progresses so does the energy it carries. Sound waves can travel through the medium of water. If you have ever put your head under water you will know that you can still hear. The molecules of the medium are being displaced about an equilibrium position. Once the wave has passed the molecules will return to this position. The correct answer is (B).

2 Question No. 2 of The wavelengths of audible sounds are 17 m to m. What are the audible frequencies, assuming velocity of sound in air is 340 m/s? Question 02 (A) 20Hz 10 khz (B) 10Hz 10 khz (C) 200Hz 2 k Hz (D) 20Hz 20 khz (E) 5x 10-5 HZ Hz The wave speed equation describes the relationship between speed, wavelength and frequency. The wave speed equation describes the relationship between speed, wavelength and frequency. The wave speed equation describes the relationship between speed, wavelength and frequency. D. Correct! Use the wave speed equation and rearrange to find frequency. The shortest wavelength will give the highest frequency. Check that you correctly rearranged the wave speed equation to find frequency. Known: Shortest wavelength, λs = m Longest wavelength, λ l = 17 m Speed of sound in air, v= 340 m/s Unknown: Frequency for shortest wavelength, f s =? Hz Frequency for longest wavelength, f l =? Hz Define: Use the wave speed equation: v= λf v Rearrange to find f: f= λ Note that frequency is inversely proportional to wavelength, so the longest wavelength should give the highest frequency. 1 Output: f = = = 20,000 s s v λs 340 m/s m Remember that s -1 is equivalent to Hz and that the prefix k stands for kilo or So f s = 20 khz v 340 m/s 1 f= l = = 20 s = 20 Hz λl 17 m Substantiate: Units are correct, sig figs are correct, Magnitude is reasonable. The correct answer is (D).

3 Question No. 3 of The figure represents a sound wave, travelling through air which of these is a correct statement. B A Question 03 (A) The parts of the wave labeled A represent regions of densely packed particles, called rarefactions. (B) The parts of the wave labeled 1 B represent regions where the particle density is low, called compressions. (C) Region A is a compression, the pressure at this point is low. (D) Distance (1) is equivalent to one wavelength. (E) Distance (2) is equivalent to one wavelength. A rarefaction is a region with a low density of particles. 2 A rarefaction is a region with a low density of particles. In a compression region there is a high density of particles and a high pressure. A complete cycle of a longitudinal wave contains one complete compression region and one complete rarefaction region. E. Correct! A complete cycle of a longitudinal wave contains one complete compression region and one complete rarefaction region. The wavelength is the distance between two consecutive compression regions or rarefaction regions. In air sound waves are longitudinal waves. For a longitudinal wave the particles in the medium vibrate parallel to the direction of travel of the wave. This results in regions where the particle density is high, a compressions and regions where the density is low, a rarefaction. The distance between two consecutive points with the highest density, is one wavelength, similarly the distance between two consecutive points with the lowest density will be one wavelength. The correct answer is (E).

4 Question No. 4 of 10 Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as 4. During a storm four people standing next to each other see a lightning strike, 10 seconds later they hear the thunder. Andy estimates the lightning struck 10 miles away. Bobby estimates that it was 3 miles away and Chris estimates it was 1 mile away. Dan says you cannot tell. Who is correct? (Hints assume the light reaches them the instant the lightning is created and 1 mile is 1.6 km). Question 04 (A) Andy (B) Bobby (C) Chris (D) Dan (E) None of the them Use the formula that relates speed, distance and time. Use the formula that relates speed, distance and time. Use the formula that relates speed, distance and time. Use the formula that relates speed, distance and time. E. Correct! Use the in formula that relates speed, distance and time. Speed = distance/time. Rearrange to find distance. This gives you an answer of 2.1 miles. A good approximation is 1 mile for every 5 s. Known: Speed of sound, 340 m/s (not given but should be known) Time between lightning and thunder, t =10 s Unknown: Distance to lightning strike, d =? m Define: The speed of light is significantly larger than the speed of sound. We can assume that the light reaches the people at the moment lightening strikes. We can use the general formula for speed to calculate the distance that sound travels in 10 s. dis tance velocity =, distance = velocity x time time The units for speed of sound are m/s but the distances given in the question are miles, we must convert from meters to miles. 1 mile = 1.6 km = 1600 m Output: Distance = 340 m/s x 10 s = 3400 m 1 mile Convert to miles x 3400 m = 2.1 miles 1600 m Note : A good approximation is 1 mile for every 5 s. Substantiate: Units are correct, sig figs are correct, Magnitude is correct, The correct answer is (E).

5 Question No. 5 of The density of a certain liquid is 1000 kg/m 3 and its bulk modulus is 2.15 x 10 9 N/m 2. Calculate the speed of sound in the liquid. Question 05 (A) 1470 m/s (B) 2.15 x 10 6 m/s (C) 6.8 x 10-4 m/s (D) 1.47 x 10 6 m/s (E) Need to know Young s modulus for the liquid. A. Correct! B Use the formula for speed of sound in a liquid, v=, where B is bulk modulus and ρ ρ is the density. Remember to take the square root. You appear to have inverted the equation. Review your use of exponents. Do not confuse the equation for liquids and solids. Known: Bulk modulus of liquid, B = 2.15 x 10 9 N/m 2. Density = 1000 kg/m 3 Unknown: Velocity of sound in the liquid, v =? m/s Define: The speed of sound in a liquid: B v= ρ Remember that 1 Newton is 1 1 kg m 2 s Output: x 10 kg m/s m 6 m v= = 2.15 x 10 = 1470 m/s kg/m s Substantiate: Units are correct, sig figs are correct, Magnitude is reasonable. The correct answer is (A).

6 Question No. 6 of Which of these is an incorrect statement about musical instruments? Question 06 (A) Musical notes are produced by when part of the instrument is set into vibration at its natural frequency. (B) The vibrations set up standing waves in the instrument. (C) In a string instrument a longitudinal wave is setup along a string. (D) The standing wave produces longitudinal waves in the surround air. (E) The frequency of the longitudinal waves is the same as that of the vibrations in the instrument. A vibration is generated in an instrument either by hitting (drums) blowing (wind instrument), plucking (strings), this causes part of the instrument to vibrate at its natural frequency. For a drum it is the skin, for the violin the strings and for a wind instrument the column of air in the tube. Remember that standing waves are generated when a string or column of air vibrates at its natural frequency. C. Correct! The wave produced on the string is transverse, the strings displacement is perpendicular to the direction of travel of the wave. The transverse vibrations of the string push on the surrounding air and produce longitudinal waves. Sound waves are longitudinal waves. The sound waves do have the same frequency as the vibrating instruments. There speed and wavelength may not be the same as the vibration in the instrument. A vibration is generated in an instrument either by hitting (drums) blowing (wind instrument), plucking (strings), this causes part of the instrument to vibrate at its natural frequency. For a drum it is the skin, for the violin the strings and for a wind instrument the column of air in the tube. Remember that standing waves are generated when a string or column of air vibrates at its natural frequency. The wave produced on the string is a transverse wave; the strings displacement is perpendicular to the direction of travel of the wave. The transverse vibrations of the string pushes on the surrounding air and produce longitudinal sound waves. The sound waves the same frequency as the vibrating instruments, but there speed and wavelength may not be the same as the vibration in the instrument. The correct answer is (C).

7 Question No. 7 of The diagram shows two modes on a vibrating string. Which of these is a correct description of each of them? (1) Question 07 (A) 1 st harmonic, (2) 3 rd overtone (B) Fundamental (2) 3 rd overtone (C) 1 st Harmonic (3) 2 nd Harmonic (D) Fundamental (2) 3 rd Harmonic (E) 1 st overtone (3) 3 rd Overtone (2) An overtone refers to how many harmonics above the fundamental frequency (1 st harmonic), so its order is always 1 less than the number of harmonics. You are correct that (1) is the fundamental. Count the number of loops you see, to find the number of harmonics. D. Correct! The 1 st harmonic is also called the fundamental, its wavelength is λ/2 (There is only half of a wave cycle on the string). (2) Is the 3 rd harmonic, its wavelength is 3λ/2. The term overtone, refers to the number of harmonics over the fundamental or 1 st harmonic, so the 1 st harmonic is not an overtone. The term overtone, refers to the number of harmonics over the fundamental or 1 st harmonic, so the 1 st harmonic is not an overtone, and the order of overtone is always one less than the order of the harmonic. (1) The 1 st harmonic is also called the fundamental; its wavelength is λ/2 (There is only half of a wave cycle on the string). (2) Is the 3 rd harmonic, its wavelength is 3λ/2. The order of the harmonic refers to the number of ½ wavelengths between each end of the string. The nth harmonic has wavelength nλ/2 An easy way to identify the order of the harmonic is to count the number of loops in this case it is 3. The correct answer is (D).

8 Question No. 8 of Two strings made from the same material are strung at the same tension. One string is 0.05m long and produces a frequency 150 times higher than that of the other, how long is the other string. Question 08 (A) m (B) 7.5 m (C) 2.7 m (D) 15 m (E) 75 m This is a very short string. Make sure you are using the correct ratio. B. Correct! To answer the question you need to use the wave speed equation and the equation for speed of sound on a string. Frequency is inversely proportional to length. Given that the tension and material are the same in each case the speeds of the wave on the string must also be the same. The ratio of lengths is therefore L high : L low = 1:150 since we know L high we can find L low The length is inversely proportional to frequency. The length is inversely proportional to frequency. This is a very long string; recheck your calculation and units. Known: Length of string with highest frequency, L h = 0.05 m Ratio of frequencies f (high)/ f (low) = 150/1 Unknown: The length of string for the low frequency, L l =? m Define: We need an equation that relates frequency and length. First use the wave speed equation: v= λf and rearrange to find v frequency : f= λ Remember that for fundamental frequency the length of the string is λ/2. λ =2L and high v v f = and f low = 2Lh 2Ll Now look at the equation for velocity of a wave on a stretched string v = T μ Where μ is the linear density measure or mass/length. Since the materials are the same in each string and the tension is the same the velocity must also be the same. We take the ration of frequencies we see that, fhigh flow L = l = 150 and we can rearrange to find L l Lh L l = L h x 150 Output: L l =0. 05 m x 150 =7.5 m Substantiate: Units are correct, sig figs are correct, Magnitude is reasonable Note: In a piano the highest frequency 150 time higher than lowest frequency, in order to do this without having extremely long pianos, the low frequency string are made with material of higher linear density. The correct answer is (B).

9 Question No. 9 of Which of these statements about beats is incorrect? Question 09 (A) Beats are produced when two sound waves interfere over time. (B) Beats occur when the two waves have similar frequencies. (C) Beats of 10Hz or less are heard as variations in pitch of the sound with time. (D) The beat frequency is given by the magnitude of f 1 f 2. (E) A beat frequency of 125 Hz cannot be heard. Beats are periodic changes in intensity as two waves that are close in frequency. Over time the waves are going in and out of phase, so sometimes are interfering constructively and sometimes destructively. Beats are the name given to the audible periodic changes in intensity. These are produced when the waves are close in frequency. C. Correct! We perceive beats as a change in intensity of the sound, i.e. how loud the sound is. Pitch is a perception of how high or low the frequency of the sound is. The pitch will be constant, an average of the two interfering waves, but the loudness of the sound will change. This is the definition of beat frequency. Our brain can really only perceive beats of frequencies of 10 Hz or less. Beats are periodic changes in intensity as two waves that are close in frequency. Over time the waves are going in and out of phase, so sometimes are interfering constructively and sometimes destructively Beats are the name given to the audible periodic changes in intensity. We perceive beats as a change in intensity of the sound, i.e. how loud the sound is. Pitch is a perception of how high or low the frequency of the sound is. The pitch will be constant, an average of the two interfering waves, but the loudness of the sound will change. The beat frequency is the magnitude of f 1 f 2. Our brain can really only perceive beats for frequencies of 10 Hz or less. The correct answer is (C).

10 Question No. 10 of A train is traveling at 120 KMPH and it blows a whistle of frequency 1000 Hz. What will be the frequency of the note heard by a stationery observer 1) If the train is approaching him? 2) If the train is moving away from him? Use a value of 340 m/s for the speed of sound in air. Question 10 (A) 1100 Hz and 910Hz (B) 910 Hz and 1100 Hz (C) Both frequencies are 1100 Hz (D) Both frequencies are 910 Hz (E) None of the above A. Correct! This is an example of the Doppler effect. The frequency of the sound will change (v ± v because the source is moving, the equation for the shift is o) f' = fo, where v is (v v s) the velocity of the wave itself, v o is the velocity of the observer in this case 0 and v s is the velocity of the source, 120 km/h. We use the top set of signs when source moves towards the observer and the bottom when it moves away. Check that you are using the correct combination of signs in the Doppler effect formula. The frequencies will change depending on whether the source is moving towards the observer or away from them. The frequencies will change depending on whether the source is moving towards the observer or away from them. Use the equation for Doppler effect to calculate the shift in frequency. Known: Frequency of sound, f 0 = 1000 Hz Velocity of train, v s = 120 km/h Velocity of observer, v o = 0 m/s Speed of sound v = 340 m/s Unknown: Frequency when train moves towards observer, f t =? Hz Frequency when train moves away from observer, f a =? Hz Define: Use equation for Doppler effect. (v+v o) (v v o) f' t=f o and f ' a = fo (v-v ) (v + v ) Output: Convert km/h to m/s. s s km 1000 m 1 h m 120 x x = = 33 m/s h 1 km 3600 s 3600 s (340 m/s + 0) f' t=1000 Hz 1100 Hz (340 m/s - 33 m/s) = (340 m/s - 0) f' a=1000 Hz 910 Hz (340 m/s + 33 m/s) = Substantiate: Units are correct, sig figs are correct, Magnitude is reasonable. The correct answer is (A).

MCAT Physics Problem Solving Drill 13: Sound

MCAT Physics Problem Solving Drill 13: Sound Question No. 1 of 10 Question 1. The wave lengths of audible sounds are 17 m to 0.017 m. Find the range of audible frequencies assuming velocity of sound in