MCAT Physics Problem Solving Drill 13: Sound

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1 MCAT Physics Problem Solving Drill 13: Sound Question No. 1 of 10 Question 1. The wave lengths of audible sounds are 17 m to m. Find the range of audible frequencies assuming velocity of sound in air as 340 m/s. (A) 20Hz - 10,000 Hz (B) 10Hz - 10,000 Hz (C) 200Hz Hz Question #01 (D) 20Hz - 20,000 Hz A. Incorrect! As v= frequency * wavelength the values cannot be in this range. B. Incorrect! As v= frequency * wavelength the values cannot be in this range. Hence, incorrect. C. Incorrect! As v= frequency * wavelength the values cannot be in this range. Hence, incorrect. D. Correct v= frequency * wavelength; f1 = V / wavelength(1) = 20 Hz and f2 = V / (wavelength(2) = 20,000Hz. This is the audible range. λ 1 = 17 m. velocity of sound in air V = 340 m/s, Frequency (f) =? f 1 = V/λ 1 = 340m/s / 17m = 20 Hz Similarly λ 2 = m, V = 340 m/s, f 2 =? f 2 = V/λ 2 = 340/0.017 = ( )/17 = = 20,000 Hz Hence Answer (D) is correct

2 Question No. 2 of 10 Question 2. A steel wire 1 m long has a mass of 1g and is with a tension of 1000N. Find the speed of propagation of the transverse wave in the string. Question #02 (A) 1000 m/s (B) 100 m/s (C) 10,000 m/s (D).0001m/s (E) None of the above A. Correct! Mass per unit length of the wire (linear density) =M/m = 1/1000Kg/m Tension applied = T = 1000 N From the velocity of transverse wave in the string = V = 1000m/s B. Incorrect! Velocity of transverse wave in the string is = V = T/M and not T/M C. Incorrect Use the formula for the speed of a transverse wave in a string. Be sure to convert units so that they match. D. Incorrect! Velocity of transverse wave in the string when substituted does not give this answer and the answer is wrong. E. Incorrect! Use the formula for the speed of a transverse wave in a string. Be sure to convert units so that they match. Length of the wire = l = 1 m Mass of the wire of length 1 m = 1 g = 1/1000 kg = 10-3 Kg Mass per unit length of the wire (linear density) =M/m = 10-3 Kg/m Tension applied = T = 1000 N Velocity of transverse wave in the string = V = T/M = (1000/10-3 ) = 10 6 = 10 3 = 1000 m/s

3 Question No. 3 of 10 Question 3. A wire of length 1 m and 20 g is stretched by a tension of 800 N. Find the fundamental frequency. (A) 1000 Hz (B) 100Hz (C) 10Hz Question #03 (D) 200Hz A. Incorrect! Here, the string represents only half of a wavelength, so the wavelength is 2m. B. Correct! Length of the wire = 1 m Mass of the wire = 0.02 Kg Mass per unit length of the wire = 0.02 Kg/m Tension in the string T = 800 N First find the speed of the waves in the string v= T/M velocity = 200m/s Then, use the wave speed equation, v=fλ Here, the string represents only half of a wavelength, so the wavelength is 2m. Solve for frequency of the fundamental, 100 Hz. C. Incorrect! First find the speed of the waves in the string v= T/M D. Incorrect First find the speed of the waves in the string v= T/M First find the speed of the wave in the wire using the formula: T 800N V = = = 200m/s m.02kg/m Length of the wire = 1 m Mass of the wire = 20 g = 0.02 Kg Mass per unit length of the wire = mass/length = 0.02 Kg/m -1 Tension in the string T = 800 N This gives a wave speed of 200 m/s. You ll also need to remember that the fundamental vibration of the wave occupies only ½ of a wavelength. Since this wire is 1m long, a whole wavelength must then be 2m. Use the wave speed equation to find the frequency. v 200m/s f = = = 100s λ 2m 1 = 100Hz

4 Question No. 4 of 10 Question 4. A tuning fork A produces 6 beats/sec with another fork B of un-known frequency. On loading B with wax 5 beats/sec are produced. If the frequency of A is 256 Hz, find the frequency of B. (A) 250 Hz (B) 251 Hz (C) 257 Hz Question #04 (D) 262 Hz A. Incorrect! N = n1 - n2 When B is not loaded with wax it may be true. But B is loaded with wax. B. Incorrect! N = n1 - n2 We do not Know that A is the greater frequency. C. Incorrect! This doesn t figure in any way with the beats formula N = n1 - n2 D. Correct! n 256 = 6 or 256 n = 6 (depending on whether B has greater frequency or less) n = 262 Hz or 250 Hz On loading B with wax, let the frequency of B be n, then N 256 = 5 or 256 n = 5; n = 261 Hz or 251 Hz. N is less than n since the frequency decreases with the increase in the load on the fork. So, 262 Hz is the frequency of B. E. Incorrect! The correct answer is provided above. Let the frequency of B be n, then Number of beats heard per second is N = n 1 n 2 Then n 256 = 6 or 256 n = 6 (depending on whether B has greater frequency Or less) n = 262 Hz or 250 Hz On loading B with wax, let the frequency of B be n, then N 256 = 5 or 256 n = 5 N = 261 Hz or 251 Hz N is less than n since the frequency decreases with the increase in the load on The fork. So, 262 Hz is the frequency of B

5 Question No. 5 of 10 Question 5. 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? (The velocity of sound in air is 330 m/s) (A) 1111 Hz and 909Hz (B) 909 Hz and 1111 Hz Question #05 (C) Both frequencies are 1111 Hz (D) Both frequencies are 909 Hz A. Correct! First change the train speed into m/s. Then use the Doppler formula with the correct signs selected. B. Incorrect! Be sure to change the train speed into m/s. C. Incorrect! Doppler effect doesn t give this result. The frequencies in the two situations cannot be the same. D. Incorrect! Doppler effect doesn t give this result. The frequencies in the two situations cannot be the same. Use the Doppler effect formula f' = f o V ± V V m V 0 S First, change the train speed into m/s, 33.3m/s. In both cases, the observer isn t moving, thus v o =0 m/s. When the train is approaching, its moving towards the observer, use the top sign of (-). 330m/s ± 0 f' = 1000Hz = 1111Hz 330m/s 33m/s When the train is going away, its moving away from the observer, use the bottom 330m/s ± 0 f' = 1000Hz = 909 Hz 330m/s + 33m/s

6 Question No. 6 of 10 Question 6. Given the threshold of human hearing is W/m 2, what is the sound level in decibels of a sound with intensity W/m 2? Question #06 (A) -20 db (B) -10 db (C) 10 db (D) 20 db A. Incorrect. The sound has higher intensity than the threshold intensity, so the sound level in db will be positive. B. Incorrect. The sound has higher intensity than the threshold intensity, so the sound level in db will be positive. C. Incorrect. This would be the correct answer if the sound had an intensity 10 times larger than the threshold intensity. D. Correct! From the equation for sound level, SL = 10 log (I/I 0 ), we can compute the sound level in decibels to be 20. See the solution below. The definition of sound level in decibels is: I SL=10log I 0. The ratio of the intensity of the sound to the threshold intensity is: I 10 W/m = = I0 10 W/m The base-10 log of this ratio is 2, so the sound level is given by: SL=10log100 =10 2 =20

7 Question No. 7 of 10 Question 7. Two materials have the same densities, but different bulk moduli. If the bulk modulus of material A is 4 times the bulk modulus of material B, what is the ratio of the velocity of sound in material A to the velocity of sound in material B? Question #07 (A) 1:2 (B) 1:1 (C) 2:1 (D) 4:1 A. Incorrect! This is the opposite of the correct ratio. Consider the relation of sound velocity to bulk modulus and density. B. Incorrect. If two materials have two different bulk moduli with the same density, the ratio of sound velocity will not be 1:1. Consider how sound velocity is related to bulk modulus. C. Correct. If the bulk modulus of material A is 4 times the bulk modulus of material B, the sound velocity of material B will be twice the sound velocity of material A. See the solution below. D. Incorrect. Although the sound velocity of material A is higher than that of material B, this is not the correct ratio. Consider the square root in the formula involving speed and bulk modulus. The sound velocity, v, of material A is related to the bulk modulus of material A, B A, and the density, ρ, as follows: B A v A =. ρ Likewise, the sound velocity of material B is as follows: B B v B=. ρ The sound velocity of material B can be written in terms of the velocity of A, because we know the bulk modulus of material B is four times that of material A. 4B ρ B ρ A A v B= =2 =2v A Thus the ratio of velocities is 1:2, so the correct answer choice is (C).

8 Question No. 8 of 10 Question 8. A concert A tone of 440 Hz is played at the same time as a tone of unknown frequency. If the two tones together exhibit a beat frequency of 5 Hz, which of the following could be unknown frequency? Question #08 (A) 430 Hz (B) 435 Hz (C) 450 Hz (D) 460 Hz A. Incorrect. Consider the definition of the beat frequency. Do not divide the difference of the two frequencies by two when computing the beat frequency. B. Correct! The difference of 440 Hz and the unknown frequency must be 5 Hz or -5 Hz, since the beat frequency is the absolute value of the difference of the two frequencies. Thus this is the correct answer choice. See the solution below for a more detailed answer. C. Incorrect. Consider the definition of the beat frequency. Do not divide the difference of the two frequencies by two when computing the beat frequency. D. Incorrect. Consider the definition of the beat frequency. Do not divide the difference of the two frequencies by four when computing the beat frequency. The beat frequency, f beat, is related to the known frequency, f A, and the unknown frequency, f B, as follows: f = f -f. beat A B Substituting the values in the problem statement, we obtain: 5= 440-f. B The only choice which satisfies this relation is 435 Hz, choice (B).

9 Question No. 9 of 10 Question 9. Two sound waves of the same wavelength but from different loudspeakers are emitted towards a point. If the two path lengths towards the point differ by half a wavelength, which of the following best describes the combination of the two waves at that point? Question #09 (A) Destructive interference (B) Constructive interference (C) In phase (D) None of the above A. Correct! If the path lengths of the two sound waves differ by a half a wavelength, then the crest of one wave will meet the trough of the other wave, and they will destructively interfere. B. Incorrect. Constructive interference occurs if the path length difference is 0 or an integer multiple of the wavelength. C. Incorrect. This is essentially the same statement as choice (B), which was shown to be wrong. D. Incorrect. One of the above choices is correct. This is a conceptual problem which doesn t involve too much mathematics. If the path lengths of the two sound waves differ by a half a wavelength, then the crest of one wave will meet the trough of the other wave, and they will destructively interfere. In other words, the two waves can be said to be out of phase. There will exist other points where the waves constructively interfere, or can be said to be in phase.

10 Question No. 10 of 10 Question 10. Suppose a sound wave of wavelength 50 cm crosses from air into water. If the speed of sound in air is 350 m/s, and the speed of sound in water is 1400 m/s, what is the wavelength of the sound in water? Question #10 (A) 25 cm (B) 50 cm (C) 1 m (D) 2 m A. Incorrect. The wavelength of sound in water will larger in water, not smaller, because of the higher sound velocity. B. Incorrect. The wavelength will be different in water, although the frequency will be the same. C. Incorrect. Although the wavelength in water will be longer than the wavelength in air, this is not the correct answer. D. Correct! First find the frequency of the wave in air; this is 700 Hz. The frequency in water must be the same as the frequency in air, so we can find the wavelength in water. This is 2 m. See the solution below. First, find the frequency of the wave in air, since we are given wavelength and sound velocity. v λ 350 m/s 0.5 m air f= = =700 Hz. air It is essential to note that the frequency is the same in both media. Next, find the wavelength in water: v 1400 m/s 700 Hz water λ water = = =2 m. f So (D) is the correct answer choice.

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

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