Physics Quiz V: Sounds and Waves

If a pendulum were placed on a planet with four times the force of gravity as that on Earth, what would be the ratio of the new period to the old period?

1/2 The equation for the period of a pendulum is . If gravity is four times larger, first you should see that the period should be smaller (since g is in the denominator). The square root however reduces the effect so instead of the period being one fourth the original it is one half the original period.

If the electronic amplification of a stethoscope increases the sound level of a particular sound by 20 dB, by what factor is the energy delivered to the ear per second increased? Select one: A. 10 B. 20 C. 100 D. 200

100dB The energy delivered to the ear per second is the power to the ear, and as long as the area of the eardrum isn't changing, power is directly proportional to intensity. The answer to the question will just be what happens to the intensity of the sound wave if the sound level is increased by 20 dB. Adding 10 dB multiplies the intensity by 10, and adding another 10 dB should multiply by 10 again. Multiplying by 10 and multiplying by 10 again is the same as multiplying by 100.

A spring with spring constant k holds a 5 kg mass and is stretched 1 m. After it is released, it passes a point 50 cm past its equilibrium point with a velocity of 4 m/s. What is the value of k?

107N/m Using conservation of energy, the total energy of the system is (1/2)kA2 , where A = 1 m. This is transformed to a mix of kinetic ((1/2)mv2) and potential energy ((1/2)kx2). Therefore

The restoring force of a spring 2 m from its equilibrium position is -10 N, what is the potential energy at this point?

10J The restoring force is F = -kx. Solving gives k = 5 N/m. This value for the spring constant can then be used to solve for the potential energy which is PE = (1/2)kx^2 = 10J

Sometimes in an enclosed body of water, a standing-wave pattern emerges in the surface waves, known as a seiche. Assuming a firm, rectangular enclosure such that the edges cannot oscillate, what will the wavelength of the fifth harmonic mode of a seiche be in a narrow, essentially one-dimensional pool with a length of 30 meters?

12m This is essentially the same as a string with both ends fixed, since the question says that the enclosure is firm in such a way that the edges cannot oscillate, so there are nodes on either end and the pool is one-dimensional. For a string with both ends fixed,

For a particular rope, it is found that the third harmonic frequency is 12 Hz. What is the fourth harmonic frequency? A. 9 Hz B. 12 Hz C. 16 Hz D. 48 Hz

16Hz Using the equation fn = nf1, we can apply n = 3 to determine a fundamental frequency of 4 Hz. Using the same formula with n = 4 gives a fourth harmonic frequency of 16 Hz. The correct answer is choice C.

Two speakers are placed in a room such that there are beats being heard every 0.25 seconds when a note of 186 Hz comes from the first speaker. What is the frequency of the wave being generated by the second speaker? A. 182 Hz B. 190 Hz C. 185.75 Hz or 186.25 Hz D. 182 Hz or 190 Hz

182 Hz or 190 Hz If beats are being heard every 0.25 seconds, that is the definition of the beat's period. Thus, the frequency is 4 Hz (4 beats per second) from the T = 1 / f. The beat frequency is calculated as the difference between the two frequencies of the competing waves is the beat frequency. Beats in this case could therefore be caused either by a sound at ± 4 Hz from 186 Hz, that is 186 - 4 = 182 Hz or 186 + 4 = 190 Hz.

A 2 m long pendulum with a ball hanging from it experiences a restoring force of 2 N when it is displaced 0.2 radians, leaving it 5 cm above the ground. What is its maximum speed in the small angle approximation?

1m/s The restoring force can be used to find the mass of the ball (1 kg), but this mass is not required to solve the problem. By using conservation of energy, mgh = 1/2mv2, gives v2 = 2gh = (2)(10)(0.05) = 1 m2/s2. Therefore, v = 1 m/s.

A certain ocean surface wave has a wavelength of 4 m and a wave height of 1 m. What is the ratio of the speed of the horizontal propagation of the wave to the average vertical speed of a buoy floating in the water?

2 First, find the horizontal propagation speed. The needed relationship is between speed and wavelength, so begin with vx = fλ (where vx is the horizontal propagation speed). Next, consider the buoy. It travels a distance in some amount of time, so begin with d = vt, or v = d/t. In one period, the buoy must travel all the way down (one wave height) and all the way back up (another wave height). Therefore the average speed of the buoy must be vy = 2h/T, where h is the wave height. The ratio will then be

If a sound wave was emitted continuously from one organ at a frequency of 150 Hz and another from another organ at 148 Hz, at what frequency would beats be heard?

2Hz The equation for beat frequency is fbeat = |f1 - f2|, so first, eliminate choices A and B, because beat frequency is an absolute value and cannot be negative. Next, the difference between 150 Hz and 148 Hz is 2 Hz.

At what wave amplitude will waves begin to break in shallow water that is 5 m deep? Select one: A. 2 m B. 3 m C. 4 m D. 5 m

2m The passage says that waves break when their height is four-fifths the depth of the water, so the wave height would have to be 4 m. The question asked about amplitude, though; wave height is measured from trough to crest, but amplitude is measured from middle to crest, so amplitude is half the wave height. Thus, the amplitude is 2 m.

How long would a pendulum with mass of 2 kg have to be to produce the same period as the Cesium atomic clock?

3 × 10-21 m The length can be calculated using the equation . Rearranging this equation yields L = g / (2πf)^2. While calculating this would be very time consuming, the one over the frequency squared component leads to the only appropriate choice being C. All other choices are too large

A block attached to the end of a horizontal spring oscillates on a frictionless surface with amplitude A. When the block is A / 2 from its equilibrium position, what fraction of its total mechanical energy is kinetic? A. 1/4 B. 1/2 C. 3/4 D. This cannot be determined without knowing the spring constant and the mass of the object

3/4 Kinetic energy is (1/2)mv2 and elastic potential energy is (1/2)kx2, where x is the distance from equilibrium. Since we do not know the speed of the block when it is at x = A/2, it's better to focus on potential energy. The potential energy at x = A/2 is (1/2)k(A/2)2 = (1/8)kA2. Since total mechanical energy is conserved, we can calculate it when it's all potential energy (i.e., when x = A). Total energy is thus (1/2)kA2, and the fraction of total energy that is potential is . If 1/4 of the total is potential energy, then 1 - 1/4 = 3/4 is kinetic energy. The answer is C.

If a dolphin detects a reflection from a fish 0.5 s after generating the sound pulse, how far away is the fish? Select one: A. 160 m B. 375 m C. 750 m D. 1500 m

375 m We use the fundamental relation d = vt, but the distance from the dolphin to the fish is half of the distance traveled by the sound. If you forgot this distinction, you would get choice C. So the distance in question is d = (1500 m/s)(0.5 s)(1/2) = 375 m. If you used the speed of sound in air, you would arrive at choice A.

If Pendulum 1 were glued onto Pendulum 3 to form a single pendulum with their combined mass and combined length, what would be the net period of the combined pendulum? Select one: A. 4.0 seconds B. 5.4 seconds C. 8.0 seconds D. 21.6 seconds

4 A. It may be tempting at first to select choice B; since you are combining two pendulums you might think the period would be the sum, but this is incorrect. Choice D also is a false lead, combining the idea of combined masses with combined length. Choice C should be ruled out since it would imply only a linear relationship to mass. Choice A is the correct answer since mass is not a factor; instead, it is a factor of the square root of length.

How much is the signal from a whale attenuated between 3 m and 300 m? A. 10 dB B. 20 dB C. 30 dB D. 40 dB

40db We start with the proportionality that intensity drops off with the square of distance. This means that between these two points the intensity will decrease by a factor of 10^4. Using our equation for decibels, this means a change of 40 dB. Forgetting the square in the proportionality makes choice B look correct.

The speed of sound in air is about 340 m/s, and the wavelength of one sound wave is 80 cm. What beat frequency would be heard if a second sound wave with frequency 420 Hz is generated simultaneously? Select one: A. 5 Hz B. 80 Hz C. 420 Hz D. 425 Hz

5 Hz The question gives us the information in order to calculate the second wave's frequency and from there, calculate the beat frequency. Using the wave equation, λf = v, we get that f = 340 / 0.8 = 425 Hz. Be careful not to fall for choice D, which is the first number you need to calculate. The beat frequency is calculated as fbeat = |f1 - f2| = 425 - 420 = 5 Hz. Thus, the correct answer is choice A.

A style of singing popular in various parts of the world, known as overtone singing, requires a single singer to produce a fundamental and a loud overtone, creating the impression of two different sounds at two different pitches. If an overtone singer sings a note at a fundamental frequency of 200 Hz and enhances the second overtone, what would the frequency of the enhanced overtone be?

600Hz First, be careful to check the passage to find out which harmonic the second overtone is! According to the second paragraph, overtones are numbered beginning with the harmonic beginning above the fundamental (that is, the second harmonic, so the first overtone is the second harmonic). This means that the second overtone is the third harmonic. Next, apply the equation fn = nf1. This suggests that f3 = (3)(200) = 600 Hz.

Three sounds have intensities I1, I2, and I3. If I2 = 4I1 and I3 = 25I2 and the intensity level of the first sound is β1 = 40 dB, what is β3? (Note: β = 10log(I/I0) where I0 = 10-12 W/m2)

60dB If I2 = 4I1 and I3 = 25I2 then I3 = 100I1. The long way to solve this problem would be to use the given equation and the value for β1 to solve for I1 and then use the equation a second time to solve for β3. A quicker way would be to remember the fact that adding 10 dB to any sound means the sound is 10 times more intense. Or conversely, every time you multiply the intensity of a sound by 10, you add 10 dB to its decibel level. Thus, I3 = 10x0I1 = (10 × 10)^I1. Multiplying the intensity by 10 twice causes the decibel level to be increased by 10 dB twice. Therefore if β1 = 40 dB, β3 = (40 + 10 + 10) dB = 60 dB.

The frequency of a wave with a wavelength of 2 nm is 4 × 1011 Hz. If the frequency doubles what is the speed of the new wave? A. 400 m/s B. 800 m/s C. 1200 m/s D. 1600 m/s

800m/s By Wave Rule #1, the speed of a wave depends on the medium and the type of wave and does not change with a change in frequency. Thus, the speed of the new wave is the same as the original, which is v = λf = (2 × 10-9)(4 × 1011) = 800 m/s.

A sound source emitting waves with frequency f is moving at one third the speed of sound in air. It is being directly followed by a detector moving at one half the speed of sound. What frequency does the detector perceive? (Note: fD = fS(v ± vD) / (v ± vS), where v is the speed of sound.)

9f/8 Since the detector is moving faster than the source while approaching, the perceived frequency will be larger than the emitted frequency. Unfortunately, this does not allow us to eliminate any answer choices. Using the equation, we look at the source and detector separately to determine the correct sign in each plus/minus. Since the detector is moving toward the source, its motion alone would cause the perceived frequency to be greater than the emitted frequency. Since its speed appears in the numerator, we need to use the "+" sign. Since the source is moving away from the detector, its motion alone would cause the perceived frequency to be less than the emitted frequency. Since its speed appears in the denominator, we need to use "+" as well. Plugging numbers in, fD = f(v + v/2) / (v + v/3) = f(1 + 1/2) / (1 + 1/3) = f(3/2) / (4/3) = 9f / 8. The correct answer is choice A. Note that it is not correct to simply use the relative velocity of v/2 - v/3.

Which of the following would make an open-ocean wave more likely to break? Select one: A. Wave height suddenly and sharply decreasing. B. A gust of high-intensity wind. C. Wavelength gradually increasing as ocean temperature changes. D. Destructive interference from a water wave moving in the opposite direction.

A gust of high-intensity wind. The passage indicates in the final paragraph that deep-water breaking occurs when the wave height is greater than one-sixth the wavelength (that is, h>λ / 6). In other words, it occurs when wave height is too big or wavelength is too small. Choices A and C are both reversed: breaking is less likely if the wave height becomes smaller or the wavelength increases. Likewise, choice D is opposite: destructive interference decreases amplitude, which means that it decreases wave height, which makes the wave less likely to break. The passage also states that breaking can occur when the wind pushes on the wave, so a gust of wind, as in choice B, would increase the chances of breaking.

What is the longest period of a sound wave that can be detected well by the diaphragm on a stethoscope?

According to the first paragraph of the passage, the diaphragm can detect high frequency sounds well, from 80 to 200 Hz (eliminating choice D). Period is the reciprocal of frequency, so the lowest frequency (80 Hz) will correspond to the longest period. In particular, T = 1 / f = 1/80. Since 1/80 is approximately 1/100, the answer is choice B (and 1/80, if calculated exactly, is exactly answer choice B).

What is the approximate factor increase in energy per unit time being delivered to listeners' ears by the singer's performance in Figure 1 at 2 seconds versus at 1 second?

At 1 second, the sound level was approximately 70 dB. At 2 seconds, the sound level was approximately 90 dB. That's 20 dB higher, which corresponds to fold increase in intensity. Since intensity is power over area, and listeners' ears remained the same area from 1 second to 2 seconds, the power (the energy per unit time) was 100 times as great.

An oceanographer measures the waves at a certain beach on a certain day and again the next day. The measurements showed that the significant wave height was twice as big on the second day as it was on the first. Which of the following must be true? Select one: A. The waves on the second day were twice as large as the waves on the first day. B. The fetch of the wind creating the waves was larger on the second day than the fetch creating the waves on the first day. C. The oceanographer measured twice as many waves on the second day as were measured on the first. D. At least a third of the waves on the second day were larger than a third of the waves on the first day

At least a third of the waves on the second day were larger than a third of the waves on the first day. Retrieve the definition of significant wave height from the passage: it is the average height of the tallest one-third of the waves. That means that it completely neglects two-thirds of the waves, so choice A may be true but does not have to be; only some of the waves need be twice as large. Choice B also could be true, but does not have to be, since the passage explains that fetch is just one of the factors that increases the energy of the waves. Choice C is possible but not necessarily true either, since significant wave height is an average, so its value can come from any number of waves counted. Choice D is true because it best fits the definition given: the tallest third on the second day had to be twice as large as the tallest third on the first day in order for the significant wave height to double.

Which of the following statements about longitudinal waves is FALSE? A. Sound waves are examples of longitudinal waves. B. In longitudinal waves, the directions of oscillation and propagation are perpendicular to each other. C. In a longitudinal wave, areas of high pressure are known as compressions and areas of low pressure are known as rarefactions. D. The speed of a longitudinal wave is dependent on the type of wave and the medium in which it is transferred.

Choices A and C are true based on the known properties of sound waves. Choice D is a property of all waves. In transverse waves, oscillation and propagation are at right angles, but in longitudinal waves they are parallel. Therefore, choice B is a false statement and is the correct answer.

Which of the following would be the effect of decreasing the space between the vocal folds as air passes through them, if the airflow is ideal and incompressible? Select one: A. Flow speed would increase, and air pressure would increase. B. Flow speed would increase, and air pressure would decrease. C. Flow speed would decrease, and air pressure would increase. D. Flow speed would decrease, and air pressure would decrease.

Flow speed would increase, and air pressure would decrease. This is a two-by-two question: knowing whether the change to either flow speed or air pressure will eliminate two answers, and knowing the change to the other will eliminate the final wrong answer. By the Continuity Equation (A1v1 = A2v2), if the available area through which to flow decreases, the speed increases to keep the flow rate constant. Eliminate choices C and D. Because of the Bernoulli Effect (flow speed increasing means pressure decreases), choice B is the answer.

If the vocal folds act like ropes on which standing waves are created, and if the frequencies of sound they create are equal to their frequencies of vibration, which of the following would be true of a man with longer vocal folds? A. He would have a louder voice. B. He would have a softer voice C. He would have a lower-pitched voice. D. He would have a higher-pitched voice.

He would have a lower-pitched voice. The equation for frequencies of harmonics of waves on ropes is (pictured). This means that frequencies of harmonics are inversely proportional to the lengths of the ropes. Increasing the length of the rope will decrease the frequencies of the harmonics, so the man would have a deeper lower-pitched (lower fundamental frequency) voice. Loudness and softness of the voice have to do with amplitude, which is unrelated to the length.

A mass connected to a spring is set in simple harmonic motion. Which of the following quantities reach their maximum value when the mass is passing through equilibrium? I. Potential energy II. Kinetic energy III. Acceleration

II only As the mass passes through equilibrium, the spring is unstretched, indicating that potential energy must be zero. Therefore, by energy balance, kinetic energy must take its maximum value at equilibrium. Similar to potential energy, spring force is also zero at equilibrium, indicating that acceleration must also be zero. Therefore, the correct answer is choice C.

Which of the following correctly matches the description of sound's longitudinal pressure wave description with the types of speakers involved? Select one: A. Midrange speakers have rarefactions and compressions closer together than tweeters. B. Midrange speakers have rarefactions and compressions farther apart than tweeters. C. Woofers have rarefactions and compressions closer together than tweeters. D. Woofers have rarefactions and compressions closer together than midrange.

Midrange speakers have rarefactions and compressions farther apart than tweeters. Recall that the definition of a longitudinal pressure wave consists of alternating rarefactions (regions of low pressure) with compressions (regions of high pressure). The closer together these rarefactions and compressions are, the shorter the wavelength and thus the higher the frequency. Recall the wave equation λf = v. The passage tells us that the order of speaker drivers in terms of increasing frequency is woofers, midrange, and then tweeters. Only choice B puts these in the correct order. The size correlation with the speaker drivers also makes sense with the frequencies produced. If one wants to create a high frequency wave, the cone must be moving very quickly, which is easier when the mass of the cone is less. (It is smaller in a tweeter compared to a woofer.)

As a swell passes from warmer water to colder water, which of the following is likely to happen? Select one: A. Frequency decreases B. Frequency remains constant. C. Frequency increases. D. The effect on frequency cannot be determined from the information given.

One of the basic rules of waves is that the frequency of a wave does not change when the wave moves from one medium to another. Here, the wave is moving from water of one temperature to water of another, which is a change in medium, so frequency should remain constant.

Energy emitted from the atomic clock travels at the speed of light in a vacuum (3 × 108 m/s). What would be the wavelength of the energy?

Passage states frequency= 9,192,631,770Hz (cycles/second) This problem can be solved by knowing that the wavelength times the frequency equals the wavespeed. Rearranging this equation yields λ = v / f. Plugging into this equation yields 3 × 108 / 9,192,631,770 which is approximately 3 × 108 / 10^10. This is equivelent to a wavelength of 0.03 meters, or 3 centimeters.

If the energy of a photon were equal to the photon's kinetic energy, what would the mass of a photon with a wavelength of 600 nm be?

Setting the equation for the energy of a photon equal to the kinetic energy and solving for the mass gives

Which of the following presents a difficulty for electronic stethoscope use? Select one: A. The amplification system may enhance the intensity of ambient sounds near the system that are not of medical interest. B. Low resistance in the wires connecting the chestpiece to the earpieces is needed for high sound fidelity but may cause currents to reach dangerously high levels, increasing the risk of electric shock for the listener. C. Low frequency sounds from the bell require more amplification to be audible, and electronic amplification is a constant that does not depend on frequency. D. Standing waves in the tube part of the stethoscope can destructively interfere with the electric signal, reducing sound volume.

The amplification system may enhance the intensity of ambient sounds near the system that are not of medical interest. Use Process of Elimination. Choice A seems possible, because the second paragraph of the passage indicates that all sounds picked up by the electronic stethoscope will be amplified, and this may make it hard to distinguish important sounds from unimportant ones. Eliminate choice B because electric shock would not be a risk unless the listener actually held the live wire through which the electric current was traveling. Eliminate choice C because there is no reason to believe, based on the passage, that electronic amplification cannot be adjusted for frequency, even if it were assumed that low frequency sounds are harder to hear. Eliminate choice D because standing waves in the tube would be sound waves, but the electric signal would be a current, and these are different types of waves; they cannot interfere.

Given that a bat uses echolocation in air and a whale in water, how will their detection times and ranges compare? A. The bat has slower detection over a shorter range. B. The bat has faster detection over a shorter range. C. The bat has slower detection over a longer range D. The bat has faster detection over a longer range

The bat has slower detection over a longer range As stated in the passage, sound travels faster in water than air, so the bat will have slower detection time. Sound is also attenuated more in water, so the bat will be able to detect things further away without the signal being absorbed.

How does the sound heard vary if there is an increase its wavelength and amplitude? Select one: A. The new sound perceived is higher pitched and softer. B. The new sound perceived is higher pitched and louder C. The new sound perceived is lower pitched and softer. D. The new sound perceived is lower pitched and louder.

The new sound perceived is lower pitched and louder. This is a two-by-two question. Recall the wave equation λf = v, indicating the inverse relationship between frequency and wavelength. Thus, if wavelength has increased, frequency has decreased, leading to a lower pitched sound being perceived (choices A and B can be eliminated). Amplitude of a wave has no effect on its frequency but instead affects the loudness. If amplitude has increased, the sound will be louder, so the correct answer is choice D.

Which of the following would allow a trained listener to differentiate between a note played by a saxophone and a note at the same pitch and intensity played by a trumpet? A. The fundamental frequencies would be different. B. The loudnesses would be different. C. The vocal folds of the players would be different. D. The overtones would be different.

The overtones would be different. Eliminate based on the second paragraph of the passage. Choice A is wrong because two notes at the same pitch should have the same fundamental frequencies. Choice B is wrong because the same pitch and intensity means the "loudness" of both instruments will be the same. Choice C is wrong because vocal folds factor into singing, but this is quite different. Choice D is right because the question specifies that pitch is the same and implies that loudness is the same, so the only factor remaining is timbre, and that is related to overtones according to the passage material from paragraph 2.

If the initial photon into the crystal is blue with a wavelength of 400 nm, what must the wavelength be of the red photons that exit the crystal?

The passage indicates that the outgoing photons have the energy of half the incoming photon. The speed stays the same, indicating that the wavelength must double, giving 800 nm = 8 × 10^-7 m

As you speed down the highway, you notice flashing lights in your rearview mirror and slow down. What change in pitch of the siren do you perceive?

The pitch increases This question deals with the Doppler effect: as a source and a detector get closer together, the detector frequency (frequency perceived by you) is higher than the source frequency. Correspondingly, as a source and detector get farther apart, the detector frequency is lower than the source frequency. When you see the flashing lights, presumably, you slow down and the cop car continues to move towards you. Thus, applying the Doppler effect principle, you'll hear an increase in frequency of the sound of the siren.

Which of the following was true of the speed of the sound graphed in Figure 1 at 2 seconds compared to the speed at 1 second? Select one: A. The speed at 2 seconds was slower than the speed at 1 second. B. The speed was approximately the same at 2 seconds as at 1 second. C. The speed at 2 seconds was greater than the speed at 1 second. D. The speeds differed enough among the fundamental and the harmonics that no generalizations are possible

The speed was approximately the same at 2 seconds as at 1 second. One of the basic rules of waves is that the speed of a wave depends on the type of wave (in this case, sound) and the medium (in this case, air) and little or nothing else. Though the volume of the sound changes from 1 second to 2 seconds, the speed of sound should remain the same, because speed does not depend on sound level.

Some headpieces also employ noise reduction technology to eliminate sounds not coming from the earpieces of the stethoscope. How might such technology work? Select one: A. Moving parts in the earpieces dampen noise not coming from the earpieces via the Doppler effect. B. The earpieces that cover the ears are made of a material in which sound travels at a very different speed than it does in air, thereby shifting the frequency out of the range that humans can hear. C. The noise reduction system in the earpieces constructively interferes with the sound from outside the earpieces, making the unwanted sound harder to hear. D. The total energy of the sound waves coming from outside the earpieces is reduced by adding wave energy out-of-phase with the unwanted sound waves.

The total energy of the sound waves coming from outside the earpieces is reduced by adding wave energy out-of-phase with the unwanted sound waves. Use Process of Elimination. Eliminate choice A because the Doppler effect shifts frequency, not intensity, so it cannot dampen sounds. Eliminate choice B because one of the fundamental rules of waves is that frequency does not change when the wave travels from one medium to another, so having a different medium between the air and the ear would not change the frequency. Eliminate choice C because constructive interference would increase the intensity of the sound, not decrease it. Choice D is correct because adding wave energy out-of-phase with the incoming sound would cause destructive interference, decreasing the amplitude and therefore the energy (and thus loudness) of the outside sound.

As light enters water, its speed is approximately 3/4 of its initial speed. Which of the following is physically possible? A. The wavelength is 4/3 of the original and the frequency stays constant. B. The wavelength is 3/4 of the original and the frequency stays constant. C. The wavelength is constant and the frequency is 3/4 of the original. D. The wavelength is constant and the frequency is 4/3 of the original.

The wavelength is 3/4 of the original and the frequency stays constant. The frequency of a wave is not affected by a change in medium, so it must stay constant, eliminating choices C and D. Then using the wave equation v = λf, it is found that the wavelength must be 3/4 of the original.

During a recording session, there is feedback through a speaker system that causes the total intensity to be 10 W/m2 for a brief moment before the sound returns to the original 30 dB. What is the change in sound level noted during the feedback?

This is a straight calculation question. Recall the equation for calculating sound level that we use to calculate the original intensity of the sound. The change in sound level is thus 130 dB - 30 dB = 100 dB.

The period T in seconds will always equal the wavelength λ in meters for a wave with a frequency f travelling at a speed of: A. f m/s B. 1 m/s C. f / λ m/s D. T m/s

To solve this problem, it is necessary to combine two equations. The first equation is that period is the reciprocal of frequency, or T = 1 / f. The other equation is that wavelength times frequency is the speed, or λf = v. Rearranging the second equation yields λ = v / f, andsetting the two equal yields 1 / f = v / f . Thus, for a velocity of 1, the result is a period with the same magnitude as the wavelength.

Why would a doctor use the bell to detect heart murmurs but the diaphragm to detect pneumonia? Select one: A. Artery vibration is usually much quieter than breathing abnormalities, and the bell helps detect the lower-energy sounds. B. Artery vibration is usually much louder than breathing abnormalities, and the diaphragm helps detect the lower-energy sounds. C. Turbulent blood flow usually creates low frequency sounds, but crackling in the lungs is usually high frequency. D. Turbulent blood flow usually creates high frequency sounds, but crackling in the lungs is usually low frequency.

Turbulent blood flow usually creates low frequency sounds, but crackling in the lungs is usually high frequency. The passage indicates that the bell and diaphragm detect different frequencies, so eliminate choices A and B. Next, the bell detects low frequency and the diaphragm high frequency, so turbulent blood flow (associated in the third paragraph with heart murmurs) must be low frequency and crackling in the lungs (associated in the third paragraph with pneumonia) must be high frequency.

A saxophone player changes notes from a high note to a lower frequency note. What has happened to wave speed and wavelength? A. Wave speed has decreased and wavelength has increased. B. Wave speed has not changed and wavelength has increased. C. Wave speed has decreased and wavelength has decreased. D. Wave speed has not changed and wavelength has decreased.

Wave speed has not changed and wavelength has increased Recall the wave speed equation λf = v , and Big Wave Rule #1: the speed of a wave is determined by the type of wave and the characteristics of the medium, not by the frequency. Given that the conducting medium (air) has not changed, the speed of the wave does not change. Thus, as frequency decreases, wavelength must increases proportionally.

o create constructive interference for sound waves, which of the following would be most important to match among the interfering waves? A. Wavelength B. Amplitude C. Polarization D. Loudness

Wavelength Constructive interference occurs when two waves are perfectly in phase. Even if they are in phase at some place, if they have different wavelengths, they are not in phase throughout their wave forms, so the waves cannot constructively interfere. Choices B and D are wrong because waves of different amplitudes or loudness can definitely constructively interfere, and choice C is wrong because sound waves do not have polarizations. Note: This question requires no information from the passage.

As you speed away in your spaceship from an imploding star, the pitch of the sound you hear from the implosion: A. increases B. decreases C. stays the same D. cannot be detected

cannot be detected If the Doppler effect was applicable, then the frequency of the sound you detect should be lower than the frequency of the sound emitted. Thus, choice B would be the correct answer except that the Doppler effect does not apply. One must remember that space is a vacuum, and sound cannot travel through a vacuum because there are no molecules to be compressed and rarefied. Therefore, choice D is the correct answer.

If the energy wave travels from a vacuum into a new medium where the wavelength decreases, the: A. frequency will increase and speed will remain the same. B. frequency will decrease and speed will remain the same. C. frequency will remain constant and speed will increase. D. frequency will remain constant and speed will decrease.

frequency will remain constant and speed will decrease. Solving this problem requires knowledge of two wave properties. The first property is that waves maintain their frequency when they change from one material to another (this eliminates choices A and B). Second, one must recognize that λf = v for any wave. Since the frequency is constant, if the wavelength decreases, the velocity must also decrease.

What are the units of Planck's constant?

kg∙m2/s. The units of energy are J = N∙m = kg∙m2/s2, the units of c are m/s, and the units of wavelength are m. Therefore, the units of h are kg∙m2/s.

What is the magnitude of the restoring torque for a pendulum with mass m and length L at an angle θ from the vertical position? A. mg cosθ B. mg sinθ C. mgL cosθ D. mgL sinθ

mgL sinθ For this problem you should be able to reduce to two answers by first remembering that torque is force times distance which eliminates choices to A and B. It is then necessary to recognize the component perpendicular to the radial direction is the sine of the angle. The correct answer is: mgL sinθ

A pipe of length L open at one end is producing sound in its nth harmonic mode with wavelength 0.8 m. A pipe with the same length but open at both ends is also producing sound in its (n + 1)th harmonic mode with wavelength 0.3 m. What is n?

n=3 The number n refers to the harmonic mode of the pipe open at one end. For this type of pipe, n must be odd. This eliminates choice A. Since there would be a node at one end of the pipe and an antinode at the other end, the length of the pipe must contain an odd number of quarter-wavelengths: L = nλn/ 4. A pipe open at both ends has antinodes at each end, so the length of the pipe must contain an integer number of half-wavelength. In this case, L = (n + 1)λn+1 / 2. Plugging in the values of the wavelengths and setting these expressions equal to each other, we get n(0.8 m) / 4 = (n + 1)(0.3 m) / 2, or 0.4n = 0.3(n + 1). Solving for n, we get n = 3.

A pinging device, producing sound at 300 Hz, is submerged in 30°C water. As the water is cooled to 4°C water (the temperature at which water is the most dense): Select one: A. the speed of the sound will increase and frequency will increase. B. the speed of the sound will decrease and frequency will increase. C. the speed of the sound will increase and frequency will remain constant. D. the speed of the sound will decrease and frequency will remain

the speed of the sound will decrease and frequency will remain This is a classic two-by-two question (each piece of information eliminates two answers). Considering the first half of the answer choices, the speed of sound will decrease with increasing density. This eliminates choices A and C. The frequency, however, will remain constant, which eliminates choice B and leaves choice D.

As a bat releases chirps in the forward direction: A. it must adjust its flight to compensate for the backward recoil. B. it must adjust its flight to compensate for the resulting air turbulence. C. the vibrational motion of small masses of air has no significant effect on mechanical motion, momentum, or energy. D. it becomes incrementally lighter with each lost phonon.

the vibrational motion of small masses of air has no significant effect on mechanical motion, momentum, or energy. Although sound is a form of energy propagation, it is only small masses of air vibrating and passing on this vibrational energy to adjacent air. The bat does not exert much force on the air and it therefore is not regarded as a collision or action-reaction scenario. Phonons are packets of acoustic energy considered in some solid state physics studies, but they are massless quanta of energy not to be confused with photons

For two waves generated in the same place in the ocean by winds with the same fetch and duration but one wind twice as fast as the other, which of the following represents the relationship between v1, the original speed of a wave generated by the slower wind, and v2, the original speed of a wave generated by the faster wind?

v1=v2 A basic rule of waves is that wave speed depends only on the type of wave (here, a water wave) and the medium (here, the same part of the ocean). The fact that one wind was twice as fast as the other makes no difference to the wave speed, so they should be equal. (The passage tells us that this will change the energy that the wave carries, but this does not relate to how fast the wave travels, just as, for example, louder sounds travel at the same speed as quieter sounds in the same medium.)

If a bat flying over a pond were to try detecting an insect below the surface, nearly the entire signal is reflected by the impedance mismatch at the pond's surface. However: A. whatever echo was detected would be the same frequency as the emitted pulse. B. because the speed of sound in water is much higher, whatever echo was detected would have a higher frequency than the emitted pulse, C. whatever echo was detected would be detected in the same time as if the insect were the same distance away in air. D. the sound striking the insect would be of the same wavelength as that emitted by the bat.

whatever echo was detected would be the same frequency as the emitted pulse. The frequency of the wave never changes. Both of the big rules of waves must be considered to rule out the other options. The little sound that does get transmitted will travel at a different speed when in water, so it will take less time than if the insect were in air, ruling out choice C. The sound will not change frequency between media, ruling out choice B. It will however, change wavelength with the change of media and velocity. The insect will experience a different wavelength (choice D is incorrect), but the sound will change back to its original wavelength and speed if it makes it back into the air.

Two ropes are attached, they have the same tension and the same length, but the second rope has four times the mass of the first rope. If a wave is started on the first rope with a wavelength of λ1, what is the wavelength when the wave gets to the second rope?

λ1 /2. If the second rope has four times the mass of the first rope, then it has linear density (mass per unit length) of four times that of the first rope. Therefore, v2 = v1 / 2. Since we know that when a wave passes from one medium to another the speed changes but the frequency does not, and we know that v = fλ, then f1 = f2 is the same as saying v1 / λ1 = v2 / λ2 = v1 / (2λ2). Therefore, λ2 = λ1 /2.