UIC PHYSICS 105 Fall 2014 Practice Final Exam. UIC Physics 105. Practice Final Exam. Fall 2014 Best if used by December 7 PROBLEM POINTS SCORE

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1 UIC Physics 105 Practice Final Exam Fall 2014 Best if used by December 7 PROBLEM POINTS SCORE Multiple Choice Problem 1 Problem 2 Problem 3 Problem 4 Problem Total 100 Page 1 of 7

2 MULTIPLE CHOICE QUESTIONS Clearly circle the letter of the best answer MCQ01 [3 points]: The diagram to the right shows the position versus time graph for a particle in simple harmonic motion. The particle s velocity is described by the function (A) cos (B) cos (C) sin (D) sin Answer (B): From graph, at t = 0, and at t = T/2 cos function describes correctly the particle s velocity as a function of time MCQ02 [2 points]: In previous question (see MSQ01), the particle s maximum speed is (A) 1.6 m/s Answer (A): (B) 3.2 m/s (C) 6.4 m/s. Form graph, A = 0.1 and T = 0.4 s 0.1 = 1.6 m/s. (D) 9.8 m/s MCQ03 [3 points]: A vertical spring has a mass hanging from it, which is displaced from the equilibrium position and begins to oscillate. At what point does the system have the least potential energy? Answer (D): (A) at the highest point kd = mg d = mg/k (B) at the lowest point, so U = Umin when y = 0 i.e. at equilibrium (C) at the point where the spring is unstretched (D) at the equilibrium point MCQ04 [2 points]: A simple pendulum consists of a 1.0-kilogram brass bob on a string about 1.0 meter long. It has a period of 2.0 seconds. The pendulum would have a period of 1.0 second if the (A) string were replaced by one about 2.0 meters long Answer (B): 2l/. l /4 (B) string were replaced by one about 0.25 meters long If T = 1 s, then l = /( ) = 0.25 m (C) bob were replaced by a 0.25-kg brass sphere (D) bob were replaced by a 4.0-kg brass sphere MCQ05 [3 points]: The figure to the right shows the potential energy (PE) diagram and the total energy (TE) line of a moving particle. If the particle s motion is periodic, what is the amplitude of this motion? (A) The motion is not periodic (B) A = 1 cm (C) A = 2 cm (D) A = 3 cm Answer (D): Yes, there are turning points at 1 cm and 7 cm and the particle oscillates between these two points A = (7 1)/2 = 3 cm. Page 2 of 7

3 MCQ06 [2 points]: A boat is moored in a fixed location, and waves make it move up and down. If the spacing between wave crests is 20 m and the speed of the waves is 5 m/s, how long does it take the boat to go from the top of a crest to the bottom of a trough? (A) 1 second (B) 2 seconds (C) 4 seconds (D) 8 seconds Answer (B): We know that: v = fλ =λ/t, hence T = λ/v. If λ = 20 m and v = 5 m/s, so T = 4 secs. The time to go from a crest to a trough is only T/2 (half a period), so it takes 2 seconds. MCQ07 [3 points]: Four waves are described by the following expressions, where distances are measured in meters and times in seconds. Which of these waves have the same period? (A) I and III, and also II and IV (B) All of them have the same period. (C) I and IV, and also II and III (D) They all have different periods. MCQ08 [2 points]: The diagrams show three identical strings that have been put under tension by suspending blocks of 5 kg each. For which is the wave speed the greatest? (A) 1 (B) 2 (C) 3 (D) 1 and 3 tie (E) 2 and 3 tie Answer (A):, cos I: 21 II: III: 21 IV: T(I) = T(III) and T(II) = T(IV) I y = 0.12 cos(3x 21t) II y = 0.15 sin(6x + 42t) III y = 0.13 cos(6x + 21t) IV y = 0.23 sin(3x 42t) Answer (D): / 1: 2: 2 /2 3: 2 2. So, in the case of 1 and 3 the wave speed is the same and greatest than in the case 2 MCQ09 [3 points]: When a wave traveling along a string goes from a heavy string to a light string, its: (A) wavelength increases, frequency decreases, speed remains the same; (B) wavelength increases, frequency increases, speed increases; (C) wavelength increases, frequency remains the same, speed increases; (D) wavelength decreases, frequency decreases, speed decreases; Answer (C): Frequency: Tension is the same in both strings. The frequency of oscillating wave in both sections is equal to the frequency at which the knot oscillates up and down, i.e. the frequency of the wave is also the same in each piece of string. Speed: /, / /,, i.e. the speed increases. Wavelength: /,, i.e. the wavelength increases. MCQ10 [2 points]: Suppose the maximum speed of a string carrying a sinusoidal wave is vs. When the displacement of a point on the string is half its maximum, the speed of the point is: (A) vs/2 (B) 2vs (C) vs/4 (D) 3vs/4 (E) Answer (E):, where and / Page 3 of 7

4 MCQ11 [2 points]: A sound wave travelling in air has a frequency f and wavelength. A second sound wave traveling in air has wavelength /2. What is the frequency of the second sound wave? (A) f/2 (B) f Answer (C): 2 (C) 2f (D) 4f MCQ12 [2 points]: Moe and Curly are standing near a speaker that is emitting a 500 Hz tone. If Moe is 2.0 m from the speaker and Curly is 6.0 m from the speaker, what is the ratio of the sound intensities heard by the two (IMoe/ICurly)? (A) 9 (B) 3 (C) 1/3 (D) 1/9 MCQ13 [3 points]: Figure to the right shows various positions of a child in motion on a swing. A monitor is blowing a whistle in front of the child. At which position, A through E, will the child hear the highest frequency for the sound of the whistle? Assume that the child is moving with SHM (simple harmonic motion). (A) A (B) B (C) C (D) D (E) E Answer (A): Intensity = power/area and since area r 2, IMoe/ICurly = (6/2) 2 = 9 Answer (C): If the child swings in SHM, the highest speed is at the equilibrium point, point C the child will hear the highest frequency at point C, and when the child is swinging toward the monitor as 1 MCQ14 [3 points]: When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the string is also clamped at its midpoint (without change in tension), the lowest four resonant frequencies are: (A) 50, 100, 150, and 200 Hz Answer (C):. (B) 50, 150, 250, and 300 Hz (C) 100, 200, 300, and 400 Hz When string is clamped at its midpoint, 2 (D) 25, 50, 75, and 100 Hz 2 the new four lowest frequencies will increase by a factor of 2 (E) 75, 150, 225, and 300 Hz MCQ15 [2 points]: If you blow across the open end of a soda bottle and produce a tone of 250 Hz (lowest possible frequency), what will be the frequency of the next harmonic heard if you blow much harder? (A) 375 Hz (B) 500 Hz (C) 625 Hz (D) 750 Hz Answer (D):, where n = 1, 3, 5, So, the frequency of the next harmonic is 3f1 = 3 250Hz = 750 Hz Page 4 of 7

5 MCQ16 [2 points]: Substance A has density of 3.0 g/cm 3 and substance B has density of 4.0 g/cm 3. In order to obtain equal masses to these two substances, the ratio of the volume of A to the volume of B will be equal to (A) 3/4 (B) 4/3 (C) 3/4 (D) 4/3 (E) 16/3 MCQ17 [3 points]: The diagram to the right shows a U-tube with cross-sectional area A and partially filled with oil of density ρ. A solid cylinder, which fits the tube tightly but can slide without friction, is placed in the right arm. The system is in equilibrium. The weight of the cylinder is: (A) ALρg (B) L 3 ρg (C) Aρ(L + h)g (D) Aρ(L h)g Answer (B): and. If then So, 4/3 Answer (A): = MCQ18 [2 points]: You put a straw into a glass of water, place your finger over the top so no air can get in or out, and then lift the straw from the liquid. You find that the straw retains some liquid. How does the air pressure P in the upper part compare to atmospheric pressure PA? (A) greater then PA Answer (C): (B) equal PA At the bottom of the straw: 0 (C) less then PA, i.e. (D) Not enough information to answer the question MCQ19 [3 points]: An object of volume V floats in water ( = kg/m 3 ) with a volume Vout above the water surface. What is the average density of a loaded barge with 20.% of its volume above the waterline? (A) 1200 kg/m 3 Answer (E): (B) 1100 kg/m 3 where is the density of water and is the (C) 1000 kg/m 3 volume below the water surface (D) 900 kg/m 3 = 800 kg/m 3 (E) 800 kg/m 3 MCQ20 [3 points]: Water (density = kg/m 3 ) flows through a horizontal tapered pipe. At the wide end its speed is 4.0 m/s. The difference in pressure between the two ends is Pa. The speed of the water at the narrow end is: Answer (E): (A) 2.6 m/s According to Bernoulli's principle (B) 3.4 m/s (C) 4.0 m/s = 5.0 m/s (D) 4.5 m/s (E) 5.0 m/s Page 5 of 7

6 SHORT PROBLEMS (You must show your work and write your answers clearly and legibly) SP1: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s. (a) [6 points] What is the mass of the block? 2 and 2 So, = 1.6 kg or. (b) [5 points] If the block oscillates with a period of 1.0 s and maximum speed of m/s. What is the maximum compression (in cm) of the spring? From conservation of energy, where the spring constant: 4 So, m = 5 cm SP2 [8 points]: A transverse wave on a string with a mass m = 8.00 g and a length l = 50 m is described by, sin Find the tension in the string. The velocity of a traveling wave in a stretched string is determined by the tension we have to find the seed of the wave. The wave speed is given by v = λ/t, where λ is the wavelength and T is the period. The displacement has the form y = ymax sin(2x/ λ + 2t/T), so 2/λ = 2.0 m 1 and 2/T = 30 rad/s. Thus v = λ/t =2λ/(2T) = (30 rad/s)/(2.0 m 1 ) = 15 m/s. Since the wave speed is given by /, where F is the tension in the string and µ = m/l is the mass per unit length. Thus = ( kg) (15 m/s) 2 /(50 m) = N SP3 [7 points]: A string fixed at both ends has successive resonances with wavelengths of 0.54 m for the mth harmonic and 0.48 m for the (m + 1)th harmonic. What is the length of the string? Let s express the length of the string in terms of the wavelength of the mth harmonic: 2 λ (1) For (m + 1)th harmonic, we can write 2 1λ (2). By solving (1) and (2) simultaneously for m: So, m L = 2.2 m.. m = 8 (λ = 0.54 m corresponds to 8th harmonic) Page 6 of 7

7 SP4: A girl is sitting near the open window of a train that is moving at a velocity of m/s to the east. The girl s uncle stands near the tracks and watches the train move away. The locomotive whistle emits sound at frequency Hz. Use = m/s as the speed of sound. (a) [4 points] If the air is still, what frequency does the uncle hear? Express your answer using four significant figures. If the observer and the source are receding from each other = Hz (b) [7 points] A wind begins to blow from the east at m/s. What frequency does the uncle now hear? Express your answer using four significant figures. Speed of sound for the observer is now = 343 m/s (10 m/s) = 353 m/s In this case, = Hz SP5: A person suffering from shortness of breath visits a doctor, who discovers that blood flow in an artery is severely restricted, noting that the flow at position B in the artery is three times faster than at position A, and is due to a plaque deposit formed at position B as shown in the diagram to the right. Assume that the heights at positions A and B are equal, the blood is incompressible, and the flow is steady and uniform. (a) [6 points] If the diameter of artery at position A is 2.0 cm, what is the diameter of artery at position B? From the equation of continuity, = constant, follows, that where,, and 3. So, = 2/1.73 = 1.16 cm (b) [7 points] If the average density of human blood is 1060 kg/m 3, and the blood s speed at position A is 0.1 m/s, determine the blood pressure change (, in Pa) in the artery between A and B? Because the height of artery is constant, the terms and will drop out We can write the Bernoulli s equation as = 42 Pa Therefore, the pressure drops by 42 Pa at the narrowest size of the artery Page 7 of 7