p250 Fa23 Exam 1 vA solutions

First Name: UMID: Last Name: Physics/BioPhysics 250 Fall 2023 - Exam #1 Form 1 SOLUTIONS READ THIS BEFORE YOU BEGIN • Print your name and UMID in the boxes provided above (please write neatly so the scanner can pick it up). • Fill in your name , UMID number , and form number (under “key”) on your scantron sheet • This is a one hour and 50 minute, closed book exam. • You may use a calculator; please do not share calculators. • The exam includes 20 multiple choice questions which will be machine graded. Each question is worth 5 points. No partial credit will be given. • You may use one 8.5×11” sheet of paper with hand-written notes written on both sides. • If you are stuck on a problem, move on and come back to it. Constants you might need ? = 8.99 × 10 9 N m 2 /C 2 𝜀 0 = 8.85 × 10 −12 C 2 /N m 2 ? = 9.8 m/s 2 Speed of sound in air at 20C = 340m/s Density of air at 20 C = 1.20 kg/m 3 Magnitude of the electron charge: ? = 1.6 × 10 −19 C Mass of the proton: ? ? = 1.67 × 10 −27 kg Mass of the electron: ? 𝑒 = 9.11 × 10 −31 kg 1 electron volt (1eV) = 1.6 × 10 −19 J Avogadro’s Number: 𝑁 𝐴 = 6.02 × 10 23 /mol Threshold of hearing 𝐼 ? = 10 −12 W/m 2 milli(m) = 10 −3 , micro( ? ) = 10 −6 , nano (n) = 10 −9 pico (p) = 10 −12 , femto(f) = 10 −15 kilo (k) = 10 3 , mega (M) = 10 6 , giga (G) = 10 9

Passage for questions 1-3: The Stethoscope A stethoscope is a medical device for listening to sounds inside the human body. It commonly consists of a chest piece that is connected to PVC tubing, which then splits into two shorter metal tubes that direct the sound to the ears of the physician. The c hest piece often has two sides: an open “bell” to better hear low-frequency sounds, and a similarly shaped hollow cavity that is covered with a flexible membrane, the diaphragm, for higher-pitched sounds. Use 𝑣 = 340 ? ? ⁄ for the speed of sound and ? = 1.2 ?? ? 3 ⁄ for the density of air for the problems below. 1. If the bell has a diameter of 25 mm on the open side that is pointing towards the patient, and the two eartips have circular openings with a diameter of 2.5 mm, by how much has the total sound intensity that reaches the physician’s ears increased, compared to what entered the bell? Assume that there are no losses in the tubing. A. It did not increase. B. By a factor of 5 C. By a factor of 10 D. By a factor of 50 E. By a factor of 1000 The power concentrated over a smaller area. 𝑃 = 𝐼𝐴 = 𝐼 𝑒?? × (2𝐴 𝑒?? ) = 𝐼 ?𝑒?? 𝐴 ?𝑒?? . Since 𝐴 ∝ ? 2 → ? 𝑒?? ? ?𝑒𝑙𝑙 = 1 2 ( 25 2.5 ) 2 = 50 . Answer D.

4. Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents the displacement. I. ? = 0.12 cos(−1.5? + 2?) II. ? = −0.12 cos(2? + 1.5?) III. ? = −0.02 cos(3? − 4?) IV. ? = 0.02 cos(−2? + 2.5?) Which of these waves have the same speed? A. I and II B. I and III C. I and IV D. II and III E. III and IV The quantity multiplying x is 2?/? and the quantity multiplying ? is 2?? . The wave speeds are 𝑣 = ?? which are 𝑣 ? = 1.33m/s, 𝑣 ?? = 0.75m/s, 𝑣 ??? = 1.33m/s, 𝑣 ?𝑉 = 1.25m/s . Answer B. 5. A phon is an experimentally derived unit of human-perceived loudness. A person perceives sound at 800 Hz to be 40 phons, which corresponds to an intensity level of 40dB (from the graph). The person hears a different sound at 100 Hz to also be 40 phons, which corresponds to 60 dB. Approximately what is the ratio of the intensity of the 800 Hz sound to the 100 Hz sound, i.e. what is 𝐼 800?𝑧 /𝐼 100?𝑧 ? A. 100 B. 0.050 C. 0.01 D. 20 E. 0.10 The intensity at 40 dB is 𝐼 800?𝑧 = 𝐼 0 10 40/10 = 10 −8 W/m 2 and the intensity at 60dB is: 𝐼 100Hz = 𝐼 0 10 60/10 = 10 −6 W/m 2 . The ratio is: 𝐼 800?𝑧 𝐼 200?𝑧 = 10 −2 Answer C. Intensity level (dB)

6. Two speakers are set up and connected to the same 1000 Hz signal source. When you stand at a particular location between the two speakers you find that the two speakers individually produce a wave with amplitude 𝐴 . When both speakers are powered, you expect: A. An amplitude which goes from 0 to 𝐴 , as the phase difference of the sound coming from the two speakers changes from 0 to 2? radians. B. An amplitude of 𝐴 when the phase difference of the sound coming from the two speakers at your position is 0. C. An amplitude of 4𝐴 when the phase difference of the sound coming from the two speakers at your position is 0 radians. D. An amplitude which goes from 0 to 2𝐴 as the phase difference of the sound coming from the two speakers changes from ? to 2? radians. E. The phase difference of the sound coming from the two speakers at your position is always zero. For destructive interference at Δ𝜙 = ±?, ±3? ± 5?, … the net amplitude is the difference which is zero. For constructive interference at Δ𝜙 = 0, ±2?, ±4?, … the net amplitude is the sum which is 2𝐴 . A: when Δ𝜙 goes from 0 to 2pi the amplitude goes from 2A to 0 and back to 2A B: When phi =0 the amplitude is 2A, not A C: The amplitude should be 2A, not 4A. (The intensity should be 4x) D: This is correct. E: There’s no way to know the phase difference without the amplitude. Answer D. 7. A guitar string plays a “G” note with a frequency of 196Hz when it is tuned to a tension of 70.0 N. A guitarist wishes to “bend” the note to a G# with a frequency of 207.65Hz. This is done by changing the tension in the string without appreciably changing its length. What string tension is required to play a G#? A. 72.1 N B. 68.0 N C. 66.1 N D. 74.2 N E. 78.6 N The frequency of a note on a closed-closed string is: ? = ? 2𝐿 = √ 𝑇 𝜇 ⋅ 1 2𝐿 → 𝑇 = (2𝐿) 2 𝜇 ? 2 ∝ ? 2 , since ? and L aren’t changing. The new tension is: 𝑇 = ( 207.65 196 ) 2 × 70N = 78.6N Answer E.

The wave will first pass upright with the peak arriving in 1.0m/2m/s = 0.5s as in C or D. The reflection will be inverted, as in C. The time for the peak of the reflection to arrive is 3.0m/(2m/s) = 1.5s Answer C. 9. An oscillator creates periodic waves on a stretched sting. The string is long enough that you don’t have to consider reflections from the end. If you halve the frequency of the oscillator, but otherwise don’t change any string properties how will the wave’s period, wavelength, and wave speed change? A. The period is halved, the wavelength is halved, and the wave speed does not change. B. The period is halved, the wavelength is doubled, and the wave speed does not change. C. The period is halved, the wavelength is halved, and the wave speed increases. D. The period is doubled, the wavelength does not change, and the wave speed doubles. E. The period is doubled, the wavelength doubled, and the wave speed does not change. The tension and mass of the string remain the same, so the wave speed doesn’t change. Halving the frequency doubles the period. And ? = 𝑣/? = 𝑣𝑇 also doubles. Answer E. 10. A duck is travelling with a speed of 0.90 m/s toward its companion, which is sitting still in the water. The moving duck paddles its feet 2.0 times per second. The paddling of the feet creates waves in the water which travel at a speed of 2.0 m/s. As the waves pass the stationary duck, it oscillates up and down. How many times per second does the stationary duck oscillate? A. 3.6 Hz B. 5.0 Hz C. 8.0 Hz D. 2.0 Hz E. 1.6 Hz The duck generating the waves is moving, so this is a moving source and stationary receiver. ? ′ = ? 0 1 1−? ? /? . Do not use the approximation because 𝑣/? is not small. With ? ? = 2.0Hz, v s = 0.90m/s , and ? = 2.0m/s we get: ? ′ = 3.6Hz . Answer A. 11. You produce a note by blowing over the hole of an open bottle. When you a blow a little harder you hear the same note but louder. Which of the following is the most likely explanation? A. As you blow harder the frequency increases, producing greater intensity for the same displacement amplitude. B. The speed with which you blow over the hole is proportional to the speed of sound. Increasing the sound speed increases the loudness. C. As you blow harder, the loudness increases through the Doppler Effect. D. As you blow harder, the amplitude of the standing waves in the bottle increase, producing greater amplitude sound waves. E. As you blow harder, sound produced by the bottle arrives in-phase at the location of the listener, producing constructive interference. Louder with the same “note” or pitch means greater amplitude but the same frequency. Answer D.

12. Sound passing though sea water is attenuated according to Beer’s Law: 𝐼 = 𝐼 0 ? −𝛽? with 𝛽 the attenuation coefficient 𝐼 the intensity, 𝐼 0 the intensity at the surface, and ? the depth. The graph shows the relative intensity ( 𝐼/𝐼 0 ) of the sound as a function of depth in sea water. Approximately what is the attenuation coefficient for this frequency sound in sea water? A. 0.0050 m -1 B. 0.10 m -1 C. 50 m -1 D. 170 m -1 E. 0.02 m -1 We need to estimate the relative intensity from the graph is about 0.80 (needn’t be exact – just choose the closest answer). Solve for 𝛽: 0.80 = ? −𝛽(50m) → 𝛽 = − 1 50m ln(0.8) = 0.0045m −1 which is much closer to A. Answer A. 13. A parent is talking to their child. To avoid hearing the parent, the child puts their ears under water. If the intensity of sound just above the water is 3.2 × 10 −5 W/m 2 , what is the intensity of sound in the water? Ignore any reduction in intensity from absorption and consider only the reflection at the surface. The acoustic impedance of air and water, respectively, are: ? ?𝑖? = 400 kg s −1 m −2 ; ? ???𝑒? = 1.48 × 10 6 kg s −1 m −2 . A. 3.5 × 10 −8 W/m 2 B. 3.2 × 10 −5 W/m 2 C. 1.1 × 10 −3 W/m 2 D. 3.2 × 10 −7 W/m 2 E. 2.5 × 10 −12 W/m 2 Transmitted sound intensity ratio is: 𝑇 ? = 4? 1 ? 2 (? 1 + ? 2 ) 2 = 1.08 × 10 −3 = 𝐼 ? 𝐼 𝑖 You could also use 𝑇 ? = 1 − 𝑅 ? . Now solve for 𝐼 ? = 𝑇 ? × 3.2 × 10 −5 W/m 2 = 3.5 × 10 −8 W/m 2 . Answer A. 1.0 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 Relative intensity 0 10 20 30 40 50 Depth (m)

A. All three charges are positive. B. All three charges are negative. C. Q 1 is positive, Q 2 is negative, Q 3 is positive. D. Q 1 is negative, Q 2 is positive, Q 3 is negative. E. Q 1 is positive, Q 2 is positive, Q 3 is negative. E points away from positives Q1 and Q2, and toward the negative, Q3. Answer E. 17. Two identical small conducting spheres are separated by 0.5 m. The spheres carry different amounts of charge, and each sphere experiences an attractive electric force of 12.8 N. The total charge on the two spheres is - 20 μC. The two spheres are now brought to touch before being separated again by 0.5 m. The electric force on each sphere is closest to which of the following? A. Zero. B. Repulsive force of 6.4 N. C. Attractive force of 6.4 N. D. Repulsive force of 3.6 N. E. Attractive force of 3.6 N. After the two identical spheres touch they will have the same charge which must be ? = −10 μC . The force on each is ? = ?? 2 ? 2 = 3.6N and it’s repulsive because they’re like charges. Answer D. 18. A proton is placed in an electric field of intensity 1 kN/C. What are the magnitude and direction of the acceleration of this proton due to this field? Given: m proton = 1.67 × 10 -27 kg, e = 1.60 × 10 -19 C. A. 9.58 × 10 9 m/s 2 opposite to the electric field. B. 9.58 × 10 9 m/s 2 in the direction of the electric field. C. 95.8 × 10 10 m/s 2 in the direction of the electric field. D. 9.58 × 10 10 m/s 2 opposite to the electric field.

E. 9.58 × 10 10 m/s 2 in the direction of the electric field. ? = ? ? = ? 𝑒 ? ? = (1.6×10 −19 𝐶)(1000N/C) 1.67×10 −27 kg = 9.58 × 10 10 m/s 2 . Positive charges experience forces in the same direction as the electric field. Answer E. 19. An electric dipole consists of charges ±q separated by 5 cm. It is placed at an angle of 45° with an electric field of intensity 30 kN/C. It experiences a torque of 5 Nm. The magnitude of the charge q on the dipole is closest to A. 1 × 10 -4 C. B. 4 × 10 -2 C. C. 6.7 × 10 -5 C. D. 4.7 mC. E. 3.9 mC. The magnitude of the torque is 𝜏 = ?? sin𝜃 with ? = ?? and ? = 5cm = 0.05m . Solving for q gives: ? = 𝜏 ?? sin45 = 4.7 × 10 −3 C Answer D. 20. A cellular membrane is composed of lipid bilayers, about 5 nm thick, which is much less than the cell radius of about 10 micrometers. We could thus consider a simplified model of a cell membrane as two infinitely large, oppositely charged parallel planes, separated by a distance d ≈ 5 nm. The charges are distributed uniformly on each plane with a charge density σ (magnitude equal but of opposite sign on both planes). 𝜖 0 is the permittivity of free space. A cation carrying a charge of +q can be at a location 1, 2, 3, 4. ? ሬԦ ? + ? − ?Ԧ 45°