093 Version TEST04 TSOI (58160) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points The gure below shows a complex wave pattern on a string moving towards a rigid hook at the wall on the right. After some time, the wave is reected from the wall. v
1
pattern traveling to the left along the string. Note: Reection about a point (hook) is the same as reection about the y -axis (wall) followed by reection about the x-axis (string). The leading part of the wave must remain in front and the wave is ipped over. 002 10.0 points Hint: Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal 2 bearing at a point O L from the lower 3 end , as shown in the gure.
Select the wave pattern for the reected wave. v 1. O
L 2L 3
v 2. correct
3. v 4. v 5.
What is the period of this pendulum? Use the small angle approximation. The moment of inertia of a uniform rod about its center of 1 mass is M L2 . 12 1. T = 2 14 L 15 g 14 L 9g 7L 12 g 2L correct 3g
Explanation: Consider the wave pattern image reected about the rigid hook on the wall. v
2. T = 2 3. T = 2 4. T = 2
v After the time it takes for the wave to be reected from the wall, this image is the wave
Version 093 TEST04 TSOI (58160) 5. T = 2 6. T = 2 7. T = 2 8. T = 2 9. T = 2 10. T = 2 62 L 105 g 38 L 63 g 19 L 24 g 43 L 72 g 26 L 21 g 26 L 45 g 1 M gL =6 1 2 ML 9 3g = . 2L
2
(3)
The expression for simple harmonic motion and Eqs. 1 and 2, yields 2 = 2M g D , IO 2 T = 2 so
IO 2M gD 1 9 1 6 M L2 M gL
Explanation: Using the parallel axis theorem, the momentum of inertia is I Md . In this case there are two masses with IO = Icm + M D , where 1 21 D= L = L , so 32 6 IO = M = 1 + 12 1 6
2 2 2
= 2
= 2 (1) keywords:
2L . 3g
(4)
L2 (2) 003 10.0 points A liquid of density 1123 kg/m3 ows with speed 1.24 m/s into a pipe of diameter 0.18 m . The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 6.29 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 atm . P1
1 M L2 . 9
When the rod is at an angle with respect to the vertical direction, the lever arm due to the weight about point O is D sin D using the small angle approximation. In turn the torque due to the weight about point O is given by the equation of motion = I , so IO d2 = m g D . dt2
D is given in Eq. 1 and IO is given in Eq. 2, and for simple-harmonic-motion, we have d2 = 2 dt2 M gD = IO
1.24 m/s 0.18 m
P2 1 atm
6.29 m
v2
0.05 m
Applying Bernoullis principle, what is the pressure P1 at the entrance end of the
Version 093 TEST04 TSOI (58160) pipe? Assume the viscosity of the uid is negligible and the uid is incompressible. The acceleration of gravity is 9.8 m/s2 and Patm = 1.013 105 Pa. 1. 334865.0 2. 1470980.0 3. 1869570.0 4. 257138.0 5. 65493.5 6. 1210180.0 7. 165459.0 8. 298369.0 9. 1385620.0 10. 176224.0 Correct answer: 1.76224 105 Pa. Explanation: Let : P2 = 1 atm . y = y2 y1 = h, since the entrance height y1 is greater than the exit height y2 . Applying Bernoullis principle to the uid ow at the entrance and exit of the pipe, P1 + g y 1 + 12 12 v1 = P2 + g y2 + v2 2 2 other end:
3
0. 86
3m
58 68
7. 38
m
P1 = P2 + g ( y 2 y 1 ) + = P2 g h +
1 [v22 v1 2 ] 2 1.013 105 Pa = (1 atm) 1 atm (1123 kg/m3 ) (9.8 m/s2 ) (6.29 m) 1 + (1123 kg/m3 ) 2 [(16.0704 m/s)2 (1.24 m/s)2 ] = 1.76224 105 Pa .
1 (v2 2 v1 2 ) 2
The lever rod is in equilibrium at angle of 68 from the vertical wall. The cable makes angle of 58 with the rod. What is the tension of the supporting cable? 1. 53.1724 2. 31.7193 3. 19.1905 4. 13.1639 5. 35.5972 6. 20.9056 7. 36.7791 8. 22.398 9. 41.4645 10. 17.1199 Correct answer: 19.1905 N. Explanation: Let : L = 7.38 m , W = 31 N , b = 0.863 m , = 68 , and = 58 .
004 10.0 points Consider a lever rod of length 7.38 m, weight 31 N and uniform density. The lever rod is pivoted on one end and is supported by a cable attached at a point 0.863 m from the
Version 093 TEST04 TSOI (58160) In equilibrium, F =0 and = 0.
4
The diagrams below show dierent standing waves on a string of length 95.2 cm. Which wave has a wavelength 27.2 cm?
Consider the free-body diagram of the lever rod: T
1.
2.
F
W
3.
4. where T is the tension of the cable and F is the force on the lever rod at the pivot point. Let the pivot point be the reference point of all torques, and let the positive direction be clockwise. Then pivot = Fpivot (0) = 0 , L grav = +W sin , 2 cable = T (L b) sin , and hence in equilibrium 0 = net = pivot + grav + cable L =0+W sin 2 T (L b) sin . Solving this equation for the cable tension T , we nd L sin T =W 2 L b sin 7.38 m sin 68 2 = (31 N) 7.38 m 0.863 m sin 58 = 19.1905 N . 005 10.0 points 8.
5. correct
6.
7.
9.
Explanation:
Let : L = 95.2 cm and = 27.2 cm . A standing wave on a string has nodes where the string does not vibrate; these nodes are spaced half-wavelengths from each other.
Version 093 TEST04 TSOI (58160) A xed end of the string cannot vibrate, so L=n 2 , n = 1, 2, 3, . . . ,
5
where n is the number of vibrating segments of the string. More precisely, the wave on the string has n antinodes and n + 1 nodes, so 2L 2 (95.2 cm) n=L = = = 8, 2 27.2 cm which means the wave has 7 antinodes and 8 nodes: one node at each end, and 6 in the middle. 006 10.0 points Earthquakes produce two kinds of seismic waves: he longitudinal primary waves (called P waves) and the transverse secondary waves (called S waves). Both S waves and P waves travel through Earths crust and mantle, but at dierent speeds; the P waves are always faster than the S waves, but their exact speeds depend on depth and location. For the purpose of this exercise, we assume the P waves speed to be 8400 m/s while the S waves travel at a slower speed of 4410 m/s. If a seismic station detects a P wave and then 95.8 s later detects an S wave, how far away is the earthquake center? 1. 481.983 2. 937.635 3. 1153.59 4. 1132.59 5. 274.988 6. 889.427 7. 1062.99 8. 337.874 9. 847.092 10. 1253.56 Correct answer: 889.427 km. Explanation: Let : vP = 8400 m/s , vS = 4410 m/s , t = 95.8 s .
S wave are both emitted at the same time t = 0, but they arrive at dierent times, red d and tS = . The S wave spectively tP = vP vS is slower, so it arrives later than the P wave, the time dierence being t = d d d (vP vS ) = . vS vP vP vS
Consequently, given this time dierence and the two waves speeds vP and vS , the earthquake center is d= vP vS t vP vS (8400 m/s) (4410 m/s) (95.8 s) 1 km = 8400 m/s 4410 m/s 1000 m
= 889.427 km away from the seismic station. 007 10.0 points This picture shows the displacements s of the air molecules in a traveling sound wave as a function of distance x.
l1 l2
Which of the following tubes, open at both ends, is closest to the right length so as to resonate at its fundamental frequency when placed in this sound wave? 1. l1 2. l2 3. l1/4 4. l2/4 5. l2/2 correct 6. l1/2 Explanation: Each end of the open tube is approximately a displacement antinode, so the length of a tube to produce a fundamental resonance 1 should be l2 . (Corresponding to the dis2 3 1 tance between l2 and l2 in the gure 4 4 above.)
and
Suppose the earthquake happens at time t = 0 at some distance d. The P wave and the
Version 093 TEST04 TSOI (58160) The velocity is 008 10.0 points Tension is maintained in a string as in the gure. The observed wave is speed 28 m/s when the suspended mass is 2.7 kg . F = m1 g
6
v= =
(2.8 kg) (9.8 m/s2 ) = 28.5138 m/s . 0.03375 kg/m
2.7 kg What is the wave speed when the suspended mass is 2.8 kg? The acceleration of gravity is 9.8 m/s2 . 1. 18.441 2. 17.4895 3. 29.0027 4. 22.2069 5. 18.1265 6. 20.0 7. 20.9874 8. 28.5138 9. 25.3598 10. 21.4373 Correct answer: 28.5138 m/s. Explanation: Let : m = 2.7 kg , v = 28 m/s , m1 = 2.8 kg , and g = 9.8 m/s2 . The tension in the string is F = mg and the mass per unit length of the string is v= = = F mg F =2 v2 v (2.7 kg) 9.8 m/s2 (28 m/s)2
keywords: 009 10.0 points A 22 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a oor with a rough surface where the coecient of static friction is 0.28 . The angle between the horizontal and the ladder is . The person wants to climb up the ladder a distance of 2.2 m along the ladder from the ladders foot.
3. 9 m =0 2. 2 m 22 kg = 0.28 Note: Figure is not to scale. What is the minimum angle min (between the horizontal and the ladder) so that the person can reach a distance of 2.2 m without having the ladder slip? The acceleration of gravity is 9.8 m/s2 . 1. 57.0894 2. 61.336 3. 62.3618 4. 66.8937 5. 69.918
= 0.03375 kg/m .
Version 093 TEST04 TSOI (58160) 6. 51.7555 7. 60.5818 8. 45.3979 9. 63.6019 10. 67.8097 Correct answer: 63.6019 . Explanation: Let : g = 9.8 m/s2 , d = 2. 2 m , L = 3. 9 m , m = 22 kg , and = 0.28 . Nw tan d L d L 2. 2 m (0.28) (3.9 m)
7
arctan arctan
63.6019 . 010 10.0 points A police car is traveling at a speed vc to the right and a truck is traveling at a speed vt to the right. The frequency of the siren on the police car is fc . vc vt Police Truck
f
P ivot
What is the frequency ft heard by an observer in the moving truck? Let vt be the speed of the observer in the truck, and vc be the speed of the source, the police car. The speed of sound in air is va . va + vt fc va + vc va vt 2. ft = fc correct va vc va vt 3. ft = fc va + vc va + vt 4. ft = fc va vc Explanation: The Doppler shifted frequency f heard in the truck is va v0 f, f = va vs where va is the speed of sound in air, vo is the speed of the observer, and vs is the speed of the source. The upper sign is used when that vehicle is moving toward the other vehicle. The relative velocity of the observer is away from the source and the relative velocity of the source is toward the observer, so va vt ft = fc . va vc 1. ft =
mg Nf
Fx : Fy : :
f Nw = 0 f = Nw Nf m g = 0 m g d cos Nw L sin = 0
Nw L sin = m g d cos f L sin = m g d cos mgd f= L tan The ladder may slip when f = fmax = Nw , so fmax m g mgd L tan
Version 093 TEST04 TSOI (58160) 011 10.0 points A block of unknown mass is attached to a spring of spring constant 7.9 N/m and undergoes simple harmonic motion with an amplitude of 4.8 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be 29.2 cm/s. Calculate the mass of the block. 1. 0.296561 2. 0.256821 3. 0.276457 4. 2.13338 5. 0.730798 6. 0.930411 7. 0.116675 8. 0.108774 9. 0.160105 10. 1.42911 Correct answer: 0.160105 kg. Explanation: Let : k = 7. 9 N /m , A = 4.8 cm , and v = 29.2 cm/s .
8
oil and salt water as shown. The block sticks up above the oil by a distance 0.16 m. The oil thickness is 0.29 m. The blocks depth in the salt water is 0.28 m. The horizontal area of the block is 0.042 m2 . The density of the oil is 819 kg/m3 , and the density of the water salt mixture is 1150 kg/m3 .
d1 d2
M
d3
Calculate the mass of the block. The buoyant force of the air is negligible. 1. 30.1488 2. 12.3318 3. 14.1202 4. 6.94988 5. 17.1941 6. 7.9833 7. 15.6407 8. 38.5366 9. 23.4994 10. 9.40848 Correct answer: 23.4994 kg. Explanation: Let : a = 0.042 m2 , d2 = 0.29 m , d3 = 0.28 m , w = 1150 kg/m3 , o = 819 kg/m3 .
If the maximum displacement (amplitude) is A A, the halfway displacement is . By energy 2 conservation, Ki + Ui = Ff + Uf 0+ A 1 1 1 k A2 = m v 2 + k 2 2 2 2 1 k A2 = m v 2 + k A2 4 2 3kA m= 4 v2 3 (7.9 N/m) (0.048 m)2 = 4 (0.292 m/s)2 = 0.160105 kg . 012 10.0 points A rectangular parallelepiped block of uniform density oats in a container which contains
2
and
The mass of the displaced salt water is m3 = w d3 a and the mass of the displaced oil is m2 = o d2 a . The mass of the block is equal to mass of the the liquid displaced, so mb = m2 + m3 = w d3 a + o d2 a = (1150 kg/m3 ) (0.28 m) (0.042 m2 ) +(819 kg/m3 ) (0.29 m) (0.042 m2 ) = 23.4994 kg .
Version 093 TEST04 TSOI (58160) 013 10.0 points A constriction in a pipe reduces its diameter from 5.9 cm to 2.4 cm . Where the pipe is wider, the uid velocity is 8 m/s . Find the uid velocity where the pipe is narrow. 1. 48.3472 2. 15.3373 3. 85.9506 4. 69.358 5. 81.2812 6. 35.1717 7. 76.644 8. 15.5204 9. 58.5744 10. 43.9453 Correct answer: 48.3472 m/s. Explanation: Let : r1 = 2.95 cm , r2 = 1.2 cm , and v1 = 8 m/s . A1 v1 = A2 v2 = constant
2 r1 v1
9
6. 1.921 7. 3.97802 8. 1.60209 9. 2.63772 10. 1.66999 Correct answer: 2.99752 Hz. Explanation: Let : lo = 0.572864 m , lc = 0.850703 m , v0 = 340 m/s .
and
The length of the open organ pipe is Lo , so the wavelength o of its fundamental should be 2Lo = 1.14573 m. For the closed organ pipe, the wavelength c of the rst overtone 4 should be Lc = 1.13427 m, where Lc is the 3 length of the closed organ pipe. Since the frequency f= v ,
the corresponding frequency of the fundamental for the open organ pipe would be fo =
2
=
2 r2 v2
v2 = v1
r1 r2
2
= (8.0 m/s)
2.95 cm 1.2 cm
v0 o 340 m/s = 1.14573 m = 296.755 Hz ,
= 48.3472 m/s . 014 10.0 points An open organ pipe of length 0.572864 m and a closed one of length 0.850703 m are sounded together. The speed of sound is 340 m/s. What beat frequency is generated by the rst overtone of the closed pipe with the fundamental of the open pipe? 1. 2.99752 2. 5.1284 3. 2.39558 4. 3.18929 5. 1.98806
while the frequency of the rst overtone for the closed organ pipe would be fc = v0 c 340 m/s = 1.13427 m = 299.752 Hz .
The beat frequency we can hear is the difference between the two frequencies above, which is then equal to Fbeats = |fc fo | = |299.752 Hz 296.755 Hz| = 2.99752 Hz .
Version 093 TEST04 TSOI (58160) 015 10.0 points A U-tube of constant cross-sectional area, open to the atmosphere, is partially lled with a heavy liquid with density 8.88 g/cm3 . A light liquid with density 2.15 g/cm3 is then poured into both arms.
10
pressure is the same as that at elevation in both columns. But Pright = Patm + h g hh + g h and Plef t = Patm + g (hh + h + h ) , so from Pascals Principle, Plef t = Pright (hh + h ) = h hh h h = 1 hh 8.88 g/cm3 1 = 2.15 g/cm3 = 2.10978 cm .
h light liquid 2.15 g/cm3 0.674 cm heavy liquid 8.88 g/cm3 If the equilibrium conguration of the tube is as shown, with a dierence in the height of the heavy liquid of 0.674 cm, determine the value of the dierence in height of the light liquid h . 1. 3.85613 2. 4.01952 3. 5.30018 4. 2.10978 5. 3.59754 6. 2.15149 7. 4.70067 8. 2.30929 9. 7.00053 10. 4.41945 Correct answer: 2.10978 cm. Explanation: Let : = 2.15 g/cm3 , h = 8.88 g/cm3 , hh = 0.674 cm .
(0.674 cm)
and
Let h be the height of the light liquid column added to the right side of the U-tube. Consider the pressure at the elevation of the light-heavy liquid interface in the left column and at point at the same elevation in the right column. By Pascals Principle, the absolute