PHYS 101 131 Exam Pack - Lecture notes all PDF

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UBC Physics Society

Physics 101 Exam Pack1 2013 c UBC Department of Physics and Astronomy and UBC Physics Society

$5.00 for Physsoc members (Exam pack only) $10.00 for non-members (Exam pack only) Composed by Sen Mei, Steven Schramm, Henry Ngo, Ray Goerke, Eric Zhao, William Scales, and Ze Fu Special thanks to Dr. John Eldridge, Dr. Fran Bates, and any other exam suppliers we may have forgotten.

For info on Exam Review Sessions, Errata, Old Posted Exams, and Exam Package Updates, please visit our website at http://physsoc.phas.ubc.ca/ Last updated: March 24, 2013 Contents: 1. Winter 2011 2. December 2010 3. December 2009 4. April 2009 5. December 2008

1 Disclaimer: This package was created with the aim to help physics students prepare for their final exam. It is by no means designed to substitute for course material or teach the course. Physsoc does not take responsibility for your performance on your exam. This package is not perfect; it may still contain minor faults and errors. We strongly encourage everyone to submit corrections and make comments on our work so that we can improve the package. Feel free to contact us by e-mail ([email protected]), or in person at Hennings 307. Please check the website for errata. Warning: Do not look into the operational end of the device; Not to be used for other purposes.

PHYS 101 Winter 2011–2012 Final

(b) A piece of glass is placed in front of the upper slit, so that the phase of the wave passing through the upper slip is changed by π radians. How far from the central point C is now the closest dark fringe? Explain your answer. (c) You set the distance between the two slits to 0.30 mm, remove the glass piece, and add another source of light (blue, with wavelength λ2 = 400 nm). Calculate the distance from point C to the closest bright spot which corresponds to a maximum for both red and blue light (excluding the one in the center at C).

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PHYS 101 Winter 2011–2012 Final Solutions

where v is the speed of sound, vs is the speed of the source, vr is the speed of the receiver, f0 is the initial frequency, and f ′ is the doppler shifted frequency. Plugging in known values, we have   340 m s−1 1244 Hz = 1200 Hz 340 m s−1 − vs −1 ∴ vs = 12.03 m s (c) Since the police car is still approaching, we can use the same formula as before:   340 m s−1 + 8 m s−1 ′ ∴ f′ = 1273 Hz f = 1200 Hz 340 m s−1 − 12 m s−1 (d) We know that the power is given by P = AI . Therefore, we can put power into the decibel formula:     P /4πr21 I1 10 log10 β1 − β2 = 10 log10 I2 P/4πr22  2   r r2 = 10 log10 22 = 20 log10 r1 r1   100 m = 6.02 dB = 20 log10 50 m Therefore β2 = 70 dB − 6 decibel = 64 dB Question 4 (20 marks) (a) dsin(θ) = (m + 12 )λ. The second dark fringe is where m = 1. 3λ for second dark fringe. 2 y 22.5 × 10−3 sin(θ) = = 5 L 600 × 10−9 3 × 5 = 200 × 10−6 = 0.2 mm. ∴d= × 2 22.5 × 10−3

d sin(θ) =

(b) 0 mm. The π phase change causes destructive interference at C. (c) Using the small angle approximation, sin(θ) ≈ θ , dy = mλ for maxima. L 600 × 10−9 × 5 = 10 mm (m = 1), 20 mm (m = 2) . . . =m 0.3 × 10−3 400 × 10−9 × 5 =m = 6.7 mm (m = 1), 13.3 mm (m = 2), 20 mm (m = 3) . . . 0.3 × 10−3

d sin(θ) = dθ = ∴ yred yblue

∴ 20 mm is the first maximum for both red and blue.

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and so we have finally that P1 − P2 − ρw g(0.2 m) g(ρHg − ρw ) 51817 m = 0.420 m = 123500

h=

Question 3a (4 marks) (i) To rank the spring constants, we begin by saying that the spring constant of one spring is equal to k . Then, recall that for springs in parallel we add the spring constants and for springs in series we add the reciprocals of the spring constants. Thus, kA = k + k = 2k 1 k kC = = 2 1/k + 1/k

kB = k kD = k

Thus, the answer is kC < kB = kD < kA . k (ii) To rank frequencies, we use the fact that ω 2 = m . We can simply compare the squares of the frequencies since this is the same as comparing the frequencies. Therefore, we have

k m k 2 ωD = m

2k m k 2 ωC = 2m

2 ωA =

2 = ωB

Hence, the answer is ωC = ωD < ωB < ωA .

Question 3b (6 marks) (i) With no damping, we have that ω0 =

r

k = 8.64 rad/s m

Hence, with damping, ωdamped =

s

k − m



b 2m

2

= 8.64 rad/s

And so the answer is given by T = 2π

r

m = 0.727 s k

(ii) To find the energy at time t = 0, recall that E0 = 21 kA20 . Thus, the answer is: 1 E0 = (56 N/m)(0.20 m)2 = 1.12 J 2 A = e−bT /2m = 0.924. So, if we multiply the previous expression And after one period, we know that A 0 for E0 by the square of this ratio, the factor of A20 will cancel and we will be left with Et=T :

Et=T = (1.12 J)(0.924)2 = 0.957 J

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Question 4 (8 marks) (a) By inspection, we can see that T = 2.0 s and so f = wavespeed is given by v = λT = 1.50m/s.

1 T

= 0.50 Hz. Also by inspection, λ = 3.0 m. Thus,

(b) To find the phase, recall that the general equation for a travelling wave is A sin(kx − ωt + φ0 ). So at (x, t) = (0 m, 0 s), this reduces to A sin(φ0 ). Hence, from the snap-shot graph, we see that D(0 m, 0 s) = m which means that φ0 = 1.03 or φ0 = π − 1.03. 0.085 m = (0.10 m) sin(φ0 ). So sin(φ0 ) = 0.085 0.10 m We have to check which value of φ0 actually works so we plug in some values for which D(x, t) ≈ 0, namely (0.5 m, 0 s):   2π (0.10 m) sin (0.5 m) + 1.03 = 0.087 6= 0 3   2π (0.10 m) sin (0.5 m) + 2.13 = −0.004 ≈ 0 3 Therefore, the answer is φ0 = 2.13 rad. (c) To produce an equation for another wave which will combine with the above wave to produce a standing wave, all we have to do is reverse the direction of travel of the wave. Recall that the direction of travel is given by the sign of ωt in the equation for the wave. Since the first wave was travelling to the right (positive x direction), ωt was negative. If we make ωt positive, we will have the desired equation:   2π x + πt + φ0 D(x, t) = (0.10 m) sin 3 In this equation, φ0 is arbitrary.

Question 5 (12 marks) (i) Recall that the equation for the Doppler shift in general is   v ± vr f ′ = f0 v ∓ vs where vr and vs are the velocities of the receiver and the source, respectively. In this case, we are def moving toward the two sources and f1 = f2 = f ′ . So, the answer is:   340 m/s + 5 m/s f ′ = (170 Hz) = 172.5 Hz 340 m/s (ii) In this case, f2 is unchanged since we are still moving toward speaker 2. But now we are moving away from speaker 1 so the new shifted frequency is:   340 m/s − 5 m/s f1 = (170 Hz) = 167.5 Hz 340 m/s (iii) At position a, we hear the same frequency from both speakers. To figure if we will hear constructive or destructive interference, we first compute the wavelength λ = vf = 2.0 m. The path length difference is ∆l = 8.0 m = 4λ so we hear constructive interference. At position b, we hear two different frequencies. Therefore, we will hear a beat with frequency f2 − f1 = 5 Hz. p (iv) At position d, ∆l = (8 m)2 + (6 m)2 − 6 m = 4 m = 2λ so we have constructive interference. (v) Consider some pointpleft of d. Call it x and suppose it is located two meters left of d. Here, δl = p (12 m)2 + (6 m)2 − (4 m)2 + (6 m)2 = 6.0 m. So we see that the difference in path length is increasing.

(vi) The maximum path length difference ∆l is 8.0 m so we will hear two interference minima when ∆l = 5 m and 7 m. 3

Question 6a—Air wedge (6 marks) (a) To calculate the number of bands, we will multiply the length by the band density. First we can calculate the relative uncertainty in the final product: 0.4 cm δNbands 1 cm + = 0.13 = 10 cm 15 cm Nbands and the actual number of bands is given by: Nbands = (10 ± 1 cm)(15.0 ± 0.4 cm) = 150 ± (150 × 0.13) = 150 ± 19 It is customary to round the uncertainty to one significant digit and report only one uncertain digit in the final result. Often this requires the use of scientific notation. Therefore, the proper way to write the answer is (15 ± 2) × 10. (b) The distance between bright bands is ∆t = λ2 . Hence, the thickness t of the spacer is given by the number of bright bands times the distance between the bands:   1 t = (150 ± 19) (516 nm) = (3.9 ± 0.5) × 10−5 m 2 Hence, the answer is t = 39 ± 5µm.

6b—Diffraction (4 marks) (i) The distance given in the figure is the distance between the two second order fringes. Hence, the diffraction angle is: 1 15.2 cm = 0.0507 rad θ2 = 2 150 cm and since a =

2λ , θ

the width of the slit is:   2 633 × 10−9 m a= = 2.5 × 10−5 m 0.0507

(ii) The relation between wavelength and diffraction angle is a sin θ = mλ This tells us that if we decrease λ, then sin θ will decrease which implies that θ will decrease as well. Hence the distance shown will be less than 15.2 cm.

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PHYS 101 Final Exam

December 2009

Above is the position vs. time plot of two simple harmonic oscillators, A (represented by the solid curve) and B (represented by the dashed curve). i) What are the amplitudes and time period for these oscillators? ii) What are the phase constants of these oscillators? iii) For each oscillator, write an equation that describes position as a function of time, evaluate all constants in the equation. iv) What is the phase difference between the tow oscillators at t = 1.0 s? v) For oscillator B, at what time(s) between 0 and 6 seconds does it have zero velocity? vi) For oscillator B, at what time(s) between 0 and 6 seconds, is it moving with maximum speed in the negative x direction?

Question 4a (8 marks) Six waves are described by the following equations. D1 = 2 sin(3x − 4t) D3 = 2 sin(−6x − 2t) D5 = 2 cos(6x − 4t)

D2 = 2 sin(4x + 3t) D4 = 2 sin(−2x + 6t) D6 = 2 sin(4x + 4t − π)

i) Which of these waves are traveling in the negative x direction? ii) Rank the waves according to their wavelengths, use equal sign when appropriate. iii) Rank the waves according to their wave speed, use equal sign when appropriate. iv) Write a equation describing the wave which when combined with wave 2 will produce a standing wave. v) Write a equation describing the wave which interferes destructively with wave 2.

Question 4b (7 marks)

2

PHYS 101 Final Exam Solutions

December 2009

December 2009 Solutions Question 2a (5 marks) All blocks are at rest, hence the net force on each block is 0. Therefore the tension in each string exactly matches the difference between forces due to buoyancy and gravity, i.e. T = Fgrav − Fbuoy

T = M g − ρwater V g

T = (M − ρwaterV )g 3

where g = 9.8 N/kg and ρwater = 1 g/cm . Note that the masses and volumes were given in grams and cm3 to simplify calculations: TA = (150 grams − 50 grams)g = 0.1 · 9.8N TB = (100 grams − 100 grams)g = 0 N

TC = (200 grams − 100 grams)g = 0.1 · 9.8N

TD = (50 grams − 50 grams)g = 0N TE = (100 grams − 50 grams)g = 0.05 · 9.8N Thus the answer is: TA = TC > TE > TB = TD

Question 2b (10 marks) i) Recall that gauge pressure is the pressure relative to the atmosphere, and that for a column of fluid at rest and points A, B in this column, PA − PB = ρg(yB − yA ) (This is just bernoulli’s equation where the velocity terms are both zero). In this case, we place point A at the water surface exposed to the atmosphere in the glass tube and point B at the bottom end of the glass tube, which is exposed to the pressure in the pipe. Note that ρwater = 1000 kg/m3 , and let us take yA = y1 to be 0, then yB − yA = h1 , hence we have gauge pressure = P1 − Patm = ρgh1 = 1000 · 9.8 · 0.45 Pa = 4.41 × 104 Pa

ii) To find v1 , recall that for fluid flow, area·velocity is constant, hence A1 v1 = A2 v2 , hence v1 = Now to find h2 , we use Bernoulli’s equation

A2 v A1 2

= 0.16 m/s.

1 1 P1 + ρv12 + ρgy1 = P2 + ρv22 + ρgy2 2 2 Note that y1 = y2 since the fluid flows horizontally, hence we can subtract the ρgy term from both sides. Furthermore, P1 = Patm + ρgh1 and P2 = Patm + ρgh2 , subtracting Patm from both sides, we get 1 1 ρgh1 + ρv12 = ρgh2 + ρv22 2 2 Divide both sides by ρ and solving for h2 , we get h2 = h1 +

1 2 (v − v22) = 0.44 m/s 2g 1

Question 3 (15 marks) i)

Amplitude of A = 0.4 m Amplitude of B = 0.4 m

Period of A = 3.0 s Period of B = 4.0 s

ii) Recall that phase constant φ appears in the generic wave equation x(t) = xmax cos(ωt + φ). For A, if we plug in t = 0, we have xmax cos(φ) = xmax , thus cos(φ) = 1, so the φ = 0. (Also φ = 2π, 4π, · · · will work, but we usually pick the phase so that 0 ≤ φ < 2π. For B, plugging in t = 0 gives xmax cos(φ) = −xmax , so that cos(φ) = −1, thus φ = π. One way to see this is that B is a cosine wave translated by half a period in time, hence the phase constant of B is half of 2π. In general, when finding phase constants you can plug in any value for t. You should try to find the one which makes solving for the phase simplest.

1

PHYS 101 Final Exam Solutions

December 2009

iii) For x(t) = xmax cos(ωt + φ), we already know xmax and φ for both A and B, we just need to find ω. Recall that ω = 2πf = 2π , where T is the period. By plugging in the values from i) and ii), we get T xA (t) = 0.4 cos( xB (t) = 0.4 cos(

2π t) m 3

2π t + π) m 4

iv) The phase difference is just the difference in the arguments of cosine, hence for t = 1.0 s, the answer is (1.0) + π] − [ 2π (1.0)] = 5π [ 2π 4 3 6 v) v = 0 ⇒

dx dt

= 0 ⇒ the x vs. t plot is “flat”, hence the answer is at t = 0 s, 2 s, 4 s

vi) This is just the t value(s) where x vs. t plot has the most negative slope, hence is answer is at t = 3 s

Question 4a (8 marks) i) Recall the generic equations for traveling waves: D = ymax sin(kx − ωt + φ) for waves traveling in the positive x direction, and D = ymax sin(kx + ωt + φ) for wave traveling in the negative x direction (to remember this, recall that sin(θ + ωt) is sin θ translated to the left by ωt). For D3 and D4 , note that sin(−θ) = sin(θ + π), for D5 , note that cos θ = sin(θ + π2 ). We now rewrite the equations as follows D1 = 2 sin(3x − 4t) D3 = 2 sin(6x + 2t + π) D5 = 2 sin(6x − 4t + π2 )

D2 = 2 sin(4x + 3t) D4 = 2 sin(2x − 6t + π) D6 = 2 sin(4x + 4t − π )

Therefore of the six equations, D2 , D3 , and D6 represent waves traveling in the negative x direction. ii) Recall k =

2π , λ

λ ∝ k1 . We have k4 < k1 < k2 = k6 < k3 = k5 , hence λ4 > λ1 > λ2 = λ6 > λ3 = λ5

ω = kω , you can remember this easily by noting the unit for ω is s−1 and the unit for k is iii) Recall v = λf = 2π k 2π −1 m , in order to get m/s, you have to divide ω by k. Doing this calculation, we find v1 = 43 , v2 = 43, v3 = 31 , v4 = 3, v5 = 32 , v6 = 1. Therefore, v4 > v1 > v6 > v2 > v5 > v3

iv) To get a standing wave, we need to combine D2 with the same wave but traveling in the opposite direction, so the equation for this wave is D = 2 sin(4x − 3t) v) For destructive interference, we need to combine D2 with the same wave but completely out of phase, this corresponds to a phase shift of π, hence the equation for this wave is D = 2 sin(4x + 3t + π )

Question 4b (7 marks) i) This is a point of constructive interference, as the distances to the two speakers are the same and the sound waves will always arrive in phase. 1 )λ so the ii) For destructive interference, ∆x, the difference in distances traveled by the two waves will be √ (n + 2 2 two waves arrive out of phase, where n is an integer greater than or equal to 0. Here ∆x= 3 + 42 − 3 = 2, and for the lowest frequency at which destructive interference happen, ∆x = λ2 . Hence λ = 2 · 2.0 = 4.0 m, m/s = 85 Hz = 340 the corresponding frequency is f = vsound 4.0 m λ

iii) Sound at 85 Hz can be restored. Due to the different distances traveled, the sound waves produced out of phase by the two speakers will arrive in phase at the chair, resulting in constructive interference when (n + 21 )λ = ∆x = 2 m, including when f = 85 Hz. However, this does not solve the problem of destructive interference for the person sitting in the chair, as destructive interference now exists at frequencies where there used to be constructive interference, i.e. when nλ = ∆x = 2 m. iv) Yes, the bench is now a point of destructive interference. Since ∆x = 0, the sound waves produced out of phase by the two speakers will always arrive out of phase.

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Phys 101 April 2009 Final

April 2009 Final Question 1 (15 marks) A vacuum lifter is a device used to lift sheets of glass. Four suction pads are attached to a frame that is lowered onto the glass. A vacuum pump reduces the pressure underneath the four pads to 0.30 atm, making a suction seal to the glass. The glass and the attached lifter can then be lifted up.

1.A) (7 marks) If the four pads each have diameters of 20.0 cm, what is the mass of the heaviest piece of glass that the lifter can carry? 1.B) (8 marks) In an effort to lift a larger load, a 2.0 m radius spherical balloon is attached to the centre of the glass. The balloon is filled with helium with a density of 0.150 kg/m3 , and the balloon when empty has a mass of 1.0 kg. The density of air is 1.20 kg/m3 . How much extra mass can the lifter carry with the help of this balloon? Question 2 (15 marks) Two masses are hanging at rest from the end of a long spring. One has a mass of 100 grams and the other a mass of 1.0 kg. The 100 g mass is abruptly cut off, and the resulting oscillations of the remaining 1.0 kg mass have an amplitude of 20.0 cm. 2.A) (4 marks) Find the force constant, k 2.B) (2 marks) Find the maximum velocity of the oscillating mass 2.C) (4 marks) If the mass reaches its maximum upward velocity (which is positive) at time t = 0, write an equation for the displacement of the mass in the form d = A cos(ωt + φ) 2.D) (5 marks) A simple pendulum is 0.30 m long. At t = 0, it is released starting at an angle of 14◦ . Ignoring friction, what will be the angular position at t = 0.65 seconds? Question 3 (15 marks) A 1.0 m long string clamped at both ends vibrates at a frequency that is three times its fundamental frequency. 3.A) (2 marks) Not counting the two ends, how many nodes does the resulting standing wave have? Draw the string displacement due to the wave. 3.B) (3 marks) If the amplitude of the string’s motion at an antinode is 2.0 cm, what is the amplitude of the string’s motion at a point exactly halfway between a node and an antinode? 1

Phys 101 April 2009 Final

3.C) (4 marks) How many nodes would a standing wave have, if it were vibrating one octave higher than in A). Ignore the two ends again. Draw the displacement pattern for this mode. 3.D) (6 marks) A 1.0 m tall glass beaker is placed under a string, vibrating in its fundamental mode. Water is slowly poured into the beaker, and the air in the beaker is heard to resonate when the water level reaches 15.0 cm and again when it reaches 71.70 cm. No resonances occur between these two points. What is the fundamental frequency of the string’s vibration? Draw the air displacement in the two modes. Question 4 (15 marks) A submarine moving towards an underwater cliff emits a sonar chirp with a frequency f = 5000 Hz. Exactly 0.80 seconds later, the reflected echo is heard, and its frequency, f ′′ is 5050 Hz. The speed of sound in water is v = 1500 m/s. 4.A) (4 marks) Write an expression for the frequency f ′ of the sound when it reaches the cliff in terms of f, v, and vsub . 4.B) (4 marks) Write an expression for the frequency f ′′ of the echo of the submarine in terms of f ′ (the frequency at the cliff), v, and vsub . 4.C) (4 marks) From the two expressions above, write an expression for the frequenc...