I have no idea how my teacher get the answers to the below questions even though I have read the explanations below for each question. Please provide complete steps and explain them in detail if possible! You don't need to rush and take your time! Thank you very much!

Question 1: You have a mass on a spring undergoing simple harmonic motion, with total energy E. You double the amplitude of the motion. What is the total energy of the system now? A) 2E B) E C) 4E + CORRECT D) E E) E The total energy stored in a mass-spring system is E = -KA?. Thus, if A is doubled, the energy is quadrupled. Question 2: The motion of a mass connected to a spring is described by x(t) = (10 cm) sin (#t), where t is in seconds. What is the first time after t = 0 when the system's kinetic energy is half of its potential energy? A) 0.478 s B) 1.91 s + CORRECT C) 1.23 s D) 0.308 s E) 4.37 s We want }mu? = } (kr?). Plugging in x(t) = Asin(wt) and v(t) = Aw cos(wt), we get... mA'w cos? (wt) = =KA? sin?(wt) 2"w? = tan?(wt) t = tan-1 (v/2) /w (The above uses the fact that w = \\k/m to cancel the w?)lliluestien 3: On Earth, a simple pendulum takes 3.00 s to complete one full swing back and forth. You take the same pendulum to the planet Remulak, and observe that it takes 1.510 s to complete the same motion. \"Fhat is the acceleration due to gravity on Hemulak? You can assume the angular displacement of the pendulum (on Earth and on Hemulak) is small. Page2of9 A} 4.1311135? 13} 0.0755 m/52 C} 17.4 mfsz {- CORRECT D} 13.1mm2 E} 5.5211135? We don't lmow the length of the pendulum, but we don't need to. The length of a simple pendulum on Earth is related to its period TE by L 2 33(3)? Here 5:3 2 9.31 nifs2 as usual. But on Remulak, that same length must. obey L = (gig-)2. Comparing these two equations, we get. _ TEE Q'syew: Plug in the numbers provided. Question 4: The motion of a piston in a car engine is simple harmonic motion. If the piston moves a maximum of 5 cm to either side of its centre point, and it has a mass of 1.5 kg, what is the maximum magnitude of the force acting on the piston when the engine is running at 4,200 revolutions per minute (rpm)? (one cycle of a car engine means one cycle of a piston). A) 33.0 N B) 4.2 kN C) 0.604 mN D) 14.5 kN + CORRECT E) 1.88 x 108 kN Here A = 5 cm 4200 1/min W = (27)f = (27 rad) . = 439.823 rad/s 60 s/min By Newton's second law, |Fmax| = mamax). For simple harmonic motion |max) = Aw. This maximum is reached when the trig function is 1 in magnitude, corresponding to the point where the piston is at the amplitude and about to change direction. And so the answer is mAw-.Question 5: The choices below provide the spring constant (k), damping constant (b), and mass (m) for the different systems consisting of a mass m undergoing damped simple harmonic motion at the end of a spring of constant k. In which systems will it take the longest for the total energy of the mass-spring system to decrease to one-quarter (1/4) of its initial value? A) (k, b, m) B) (4k, b, 2m) C) (k/2, 6b, 2m) D) (k, b, 10m) + CORRECT E) (3k, 2b, m) The total energy drops off like exp (-# ), so the smallest b/m will take longest to decay. Question 6: A sinusoidal force of amplitude A is applied to an oscillator. To maintain the largest possible oscillation amplitude, the frequency of this applied force should be A) half the natural frequency of the oscillator B) the same as the natural frequency of the oscillator + CORRECT C) twice the natural frequency of the oscillator D) irrelevant since that will not affect the amplitude E) twice the damping constant Fo/m A = V ( w2 - w ) 2 + ( bu ) 2 For max A we need w = wo.Question 7: The displacement of a string carrying a travelling sinusoidal wave is given by y(x, t) = ym sin(kx - wt - $) . At time t = 0, the element of rope at a = 0 has a velocity of zero (0) and a positive displacement. The phase constant o is: A) 00 Page 4 of 9 B) 180 C) 270' + CORRECT D) 90 E) 450 Vy (x = 0, t = 0) = -W ym cos(-) = 0 ( =90" or 270 y(x = 0, t = 0) = ym sin(-6) >0 sin(-909) = -1 sin(-2709) = +1Question 3: A pulse is travelling down a rope with a cross section of radius 7'1 which is attached to another rope of cross section radius T2 > T1. If both ropes are made of the same material, when the pulse reaches the boundary between the two ropes A} it will 1;:a.rtl:,r reect back onto rope I noninverted, and partly be transmitted onto rope 2 inverted. B} it will be transmitted as is onto rope 2. C} it will partly reflect back onto rope 1 inverted: and partly be transmitted onto rope 2 non inverted. i CORRECT D} it will partly reect back onto rope 1 inverted, and part1}r be transmitted onto rope 2 inverted. E} it will partly reect back onto rope I non-inverted, and partly be transmitted onto rope 2 noninverted. Since it's from lighter to heavier: the reection will he inverted, but some will be transmitted to rope 2 and that will he noninverted. Question 9: By what factor is the intensity of a sound at a rock concert higher than that of a whisper when the two sound levels are 120 dB and 20 dB, respectively? A) 1010 + CORRECT B) 100 C) 10 Page 5 of 9 D) 10100 E) 6 The sound levels for the concert and whisper are 120 = B. = 10log ,(le/lo) and 20 = Bw = 10 log 10(Iw/ lo), respectively. We have 100 = 120-20 = Be-By = 10log10(Ic/lo)-10 log10(Iw/ 10) = 10log10(Ic/Iw), and hence Ic/ Iw = 1010: the concert is a factor 1010 times larger in intensity than the whisper.Question 10: A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the difference (in magnitude) between the frequency that the driver of the moving car hears before passing the police car, and the frequency the driver hears after passing it? (The speed of sound in air is 343 m/s.) A) 559 Hz B) 105 Hz + CORRECT C) 448 Hz D) 453 Hz E) 552 Hz The frequency is Doppler-shifted by the motion of the car. The police car is stationary, so us = 0. As the car is approaching the police car, we take vo = +36m/s and the observed frequency is ftowards = (343 + 36)/343(500) = 552.478 Hz. As the car is moving away from the police car, we take vo = -36 m/s and the observed frequency is faway = (343 - 36)/343(500) = 447.522 Hz. The frequency difference is | ftowards - faway| = (552.478 - 447.522) Hz.Question 11: The superposition of two waves, 31(x, t) = (2 x 10-8 m) sin * ( -170#) and yz(r, t) = (2 x 10-8 m) sin * (5 -170t - NI where r is in metres and t is in seconds, results in a wave with a wavelength of A) (47 ) m B) (x) m Page 6 of 9 C) (2) m D) (7/2) m E) (4) m + CORRECT The two waves can be added, by the principle of superposition. Since each component wave has the same angular frequency w = 170% rad/s and wavenumber k = #/2 rad/m, the sum y1 + yz will also have this wavenumber. Thus A = 2x/k = 2x/(W/2) =4 m.Question 12: In a standing wave, not necessarily at the fundamental frequency, en a string cf length L, the distance between nudes is A} L/E! a} A C} M2 CORRECT 13} M4 E} L/4 A standing wave is proportional to sinUu' + 46], and the nodes are given by sinUcz + (b) = ID. Since sin[X) = 0 has solutions X = 0,11'121'1', etc... the distance between nodes is :IT radians, and hence Trfk in H1. This is 171/9an)- = ill/'2. The string length is irrelevant. Question 13: A particular vibrating string of length 53 cm on a piano forms a standing wave which has 3 antinodes. What is the wavelength of this standing wave? A) 17.712111 B} 53.012111 C) 35.312111 {- CORRECT D) 26.512111 E} 25.512111 Page?an A piano string has nodes at both ends1 and there will be 2 antinndes per wavelength such that L = (# of antinodes}- % 2L 2- 53 3 A = (# of antinndes) = 35.3333 cm Question 14: When 2 identical guitar strings (same length, same material) are strung, they each produce a note at 350 Hz. After the tension in one of the two strings is increased slightly, you can now hear 3 beats every second when the two strings are strung together. What is the frequency produced by the new tightened string? A) 347 Hz B) 174 Hz C) 353 Hz + CORRECT D) 176 Hz E) I need to know the length of the string to answer this question. If the tension is increased only slightly, then the wave speed increases while the wavelength will remain the same so the frequency will increase instead, causing the beats. As such, the new frequency will be higher than the original frequency and we have fbeat = fnew - foriginal fnew = fbeat + foriginal = 3 Hz + 350 Hz = 353 HzQuestion 15: Two objects, m1 = 700 kg and m; = 300 kg. are separated by a distance of 5 In. At what minimal distance between these 2 objects can a third object. be placed so that this third object experiences a net gravitational force of zero? A} 2.14 m from 1111 B} 3.50 m from 1111 C} 0.939 n1 from m] D} The distance cannot be determined Without knowing the mass of the third object. E} 3.02 m from 1111 { CORRECT PageBon We need the force from Inch object to balance out when pulling object 3 in opposite direction. Thus we have Gmlma _ Gmma r2 _ (5 7')\" = "LE :3 (5 7')\" mgr=m1[(5}22-(5)-r+r2] : (3' (5)\""1133: [2'(5}""11]2 -4' ("11 -m2}' [(5)2'm1] 2-0111 m2} 1' r = 3.021?3 m away from ml or (5 r) n1 from 1112