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hw_6_solu - Hawaii/aw k 6 f6{04than 5.7 A siiicon...

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Unformatted text preview: Hawaii/aw k 6 f6 {04than 5.7) A siiicon cantiiever beam is 300 um long, 100 um wide and 6 pm thick. Silicon’s modulus of elasticity is 110 GPa; its density is 2330 kg/m3. (a) Determine the spring constant, k, of this beam. (b) Determine the natural frequency in radians per second. (c) Express your answer from part (b) in hertz. ANSWER A) 6 22, M/m k 13th "(110x109Pa X100x10‘tm Xexio m )_ N, “W 4L3 4(500x10'6m W B) The mass of the beam is: m = pV = (2330kg/m3 1300x10-6m X100 x10‘6m I6x10‘6m )= 4.2 ><10'10 kg Natural frequency, can, using the effective mass of the beam, m3f024m, is: cos-seem 22W w,=—k—,/ iam- 024m 0—(.—)2442x10'i°kg :1? 65mm ramp/J (10" . . _ 2521mm 79%! kHz C) fn= 5.13) A cantilever beam has length, 1, width, w, thickness, t, modulus of elasticity, EM, and density, p. (a) Wha is its natural frequency, in Hz? (b) What is its quality factor? (0) Calculate the beam’s natur 1 frequency using the following parameters: 1 “—W 100 um, w 2 40 pm, I: 5 pm, E=100 GPa’ pWZOOO ng/mB, FOWlO nN. (d) A periodically varying driving force acts on the beam. Plot the amplitude, A, of the beam’s oscillation as a function of the drive frequency, f . The damping coefficient, b=l 0‘7 N-s/rn. Include}:2 on the graph so that the resonance peak visible. (Don’t forget, for a cantilever beam may: . 0.24m.) (c) On the same graph, piot th amplitude if the damping coefficient doubles. ”WWWMwwwmi. . at ive alu f 5.13) A cantilever beam has length, l “—” 100 pm, width, w = 40 pm, thickness, at3 5 pm, moduius of elasticity, EM=l00 GPa, density, p=2,000 kg/m3,.and damping coefficient, b=10'7 N-s/m. (a) What is the beam’s damped natural frequency, fig (Hz)? (Don’t forget, for a caniilever beam mejffl 0.24m in the cciuation for damped natural frequency.) (b) What is the quality factor? (0) A periodically varying driving force with F 0”“) nN acts on the beam. Plot the amplitude, A, of the beam’s oscillation as a function of the drive frequency, firm-W. Include]; on the graph so that the resonance peak is visible. ((1) On the same graph, plot the amplitude if the damping coefi‘icient quadruples. Label this curve “extra damping.” (c) On the same graph, plot the amplitude if, instead, the effective mass of the beam increases by 500 femtograms. Label this curve “extra mass.” (1) Finally, on the same graph, plot the amplitude if, instead, the spring constant increases by 5 N/m. AN SWER [(IOOXIOQN/mik0x104fix104311)} (1 04:4 -s/mf_ 4 (mono-5:11)} _O.9—6TIOO><10’°m 40x10‘6m 5xic"6n§)(fi)00kg/mfii - (0.24ii100x10‘6mi40x 10"6 mi5x10~tmi2000kg1m3l _ i lZSN/m —~2.6xlO“‘NI’m_- 271' 9.6x10"2kg 46 fit“ 574,300 Hz :2 HIM/M Eth3 k [3 (plwr)(0 1'1) 194%. : ling?” {C B) Q~—_____“km_ 4 M f z [(10()x105‘1\1m2 K40x10”6mX5x1(}—6fl): 40 00 10—6 )3 ]((}00x10"6m)(40x10"6mx5><10”6mX2000kg/m3))(o,2‘{) X In if 10"7N.s/m :1 car-3%. 5;: C’ D? EaF) 3.5-08 part (9) extra mass pan (0) 9611 (f) exira st'fi‘fness 115-03 SEE—08 TIE—08 - amplitude, A {in} pan {:0 1.508 extra damping ESE-09 mama . ______r f 1 W 550000 555000 560000 565000 SYGUDG 575000 580000 585000 590000 595000 600000 drive fiequency, farm [82} 47 3.508 - part (f) extra mass part (6) part (9) extra sfifiness 3.E-05 J 2.E-OB -l E <' par? (8) o E 2.E—OB « extra damping “a. E m 1.5—08 J 512-09 J 0. E+00 550000 555000 550000 565000 570000 575000 580000 535000 590000 595000 600000 drive frequency, rd [Hz] 5.14) A cantilever beam has a spring constant, #80 N/rn and an effective mass, mef7.68><10'11kg. The coefficient of damping is b=10‘5 N-s/m. (a) Determine the quality (Q) factor of this system. (b) Plot the ampiitude, A, of the beam’s osciliation as a function of the drive frequency, )2; for a driving force, F93300 nN. Estimate Q for this system using your graph—labeling the resonance frequency, fin and the bandwidth, Af, you measure. ANSWER —11 A)Q_ J—_ 1/(.801s1/mi763x10 kg)=z— 0‘5 Ns/m B) 4.508 3.5438 3.503 @, A =2.ueee m, '5 2503 M: 20,850 32 ¢ a“ ‘u g i g‘ 2 5—08 :6 1.508 3 secs 4 fn=€62,43? 0.E+00 WWWWMWWW : fi—mm— “1W 1 60000 80000 100000 120000 140000 360000 180000 200000 220000 240000 drive frequency, fd [Hz] The resonance frequency on the graph is 162,437 Hz. The maximum amplitude on this graph is 2.94X 10'8 n1, so the bandwidth is measured between the points where the amplitude equals: 2.94 x108 m 45 The bandwidth Af=20,850 Hz. m2.08x10'3m f" #162437 W 73 Qfi’Af 20850 ‘— 5.16) The Lennard—Jones potential energy between a pair of atoms is determined to be: )m 2.3x10“1341°m12 _ ataxia—“1m“ 2 x1 x6 PE(x (a) Plot the potential energy curve as a function of separation distance and determine the equilibrium separation, x9. (‘0) Determine the force between these two atoms at x9. (c) What is the spring constant, 76, of this bond? (d) What is the natural frequency, expressed in hertz, of this atomic pair iftheir masses are 4.12x10‘26 kg and 2.78><10‘26 kg? ANSWER A) x3 m 0.3 nm {see graph! —2E—20 ~4E-20 Xe=0.3 nm -6EE—2O Off-+00 2.5540 4.540 6.510 3.510 “LE-09 separation, x (m) B) The force is the derivative the energy with respect to x. A»? .W €577“); LEAVHAW‘ ) 27.6x10”‘34J-.m12 39.6x1o-771-m6 f‘Df'fiow/ Hm, =WWWW+m——-7—-— “RIF“ EJ‘ «2% x x 4y £33m He‘s-fl ‘ F (x dx (cane-9111)“ (0.3x10“9m) D) We first detetmine the reduced mass, my: mgn2 _ (4.i2><10'26kg .78x10'25kg) ~166x10'26kg m1+m2 4.12x10’26kga— .78x10“26kg ' m,= The natural frequency is: 1 k ”3.... 33N/m 2a: m 2:: 1.66x10‘26kg r 2 7.1><1012 Hz, or about 7 T1125 5.21) A hannonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon. (a) What is the energy of this photon in terms of k and m? (b) If the oscillator is a bonded atom with k=15 N/rn and m~—~~4t><10'26 kg, what is the frequency (Hz) of the emitted photon? (Note: the energy of a photon, Ephom = hf) (c) In yvhich region of the electromagnetic spectrum (may, visible, microwave, etc.) does this photon belong? ANSWER A) Each quanta of energy is separated from the next by hf”. Therefore 3 quanta would be 3hfl,. This is given by: sizffifl 5 2am B) The photon has an energy of: —34 _ fifi=36626x10 I s) 1513? m6.12x10'21.l 27: m 27: 4x10 kg Its firequency is therefore: E -21 f = ”m" -~————6'12xm I m9.25><10”Hz h " 6.626x10‘34 J-s C) This is in the infrared re ion of the s ectrum. 5.27) A chemistry student accidentaily drops a large mercury thermometer and it breaks. The thermometer contained 2 grams of mercury (Hg), a potent neurotoxin. The Hg leaks out and pools together into a droplet on the floor of the laboratory. The Occupational Safety and Health Administration (OSHA) sets the permissible exposure limit (PEL) for mercury vapor in air at 12 parts per billion, by volume. At no time should anyone be exposed to mercury vapor above this threshold. The molar mass of Hg is 201 g/mol. The temperature and pressure in the lab are 14.7 psia and 70°F, respectively. (a) Convert the Hg vapor PEL to mg/rn3. (b) A professor immediately evacuates the room and seals it off so that no air can flow in or out. The drop begins to evaporate from the drop into the air at a rate of 0.4 rig/min. The laboratory (air) measures 100 m3. Within 45 minutes a cleanup crew arrives. Has the mercury vapor concentration exceeded the PEL by this time? (We assume an even distribution of the evaporated mercury throughout the lab.) (0) The cleanup crew has a cantilever mass sensor. The cantilever is coated in gold, which binds selectively to mercury vapor. The cantilever is 655 um long, 10 um Wide, with a modulus of elasticity, EM: 100 GPa, and density, pmz330 kg/m3 and the minimum detectable frequency change, AfmumOd Hz. What is the minimum detectable mass of this sensor? (d) What’s the minimum number of Hg molecules that must collect on the sensor in order to equal the minimum detectable mass? (Assume that only Hg collects on the sensor.) (e) The sensor system is coupled to a pump that pulls in a 10 his sample of the contaminated air (after the 45 minutes of Hg evaporation). What is the minimum percentage of the mass of mercury in this sample that must be collected on the cantilever sensor in order to detect the mercury? ANSWER (0.012ppm X201g/mole4.7psia)é 3 A c 3 2m_v.____._____ $0.1 g/ ) [mg/m1 0.6699(459.7+70F) m m B) The concentration is: 5 . . W m l.8x10'7 g/tn3 No. This does not exceed the PEL. 100m ___________.________ 3 3 l2.__________L__ C) mm"- prAfmm 330kg/m )(655x10 mMOxio m 01s )1'0x1017kg, W/EM 100x109Pa or 100 femtograms D) First we determine the mass of one Hg atom: 201 g/mol a ——23—- w 3.34x10'22 g/atom 6.022 x 10 atoms/mol So the number of atoms needed to equal the minimum detectable mass is: 100x10"15 g *~——_2§— =309 million atoms 3.34 x10 g/atom E) The mass of Hg contained in this volume is given by: m .—. [mg/m3]. IV): (1.8 x10”? g/tn3 110313 )m 1.8 ><10‘6 g The sensor therefore needs to collect the following percentage: mm _100x10'”g 1.8x10~6 g a 5.6 x 10-3, or 5.6x10" Qercent m sample 5.39) A chemical adsorbs onto the surface of a rectangular silicon beam measuring 200 um long and 1 pm thick, with an elastic modulus of 100 GPa and a Poisson’s ratio of 0.22. This alters the surface stress by 2 X 10‘4 N/m. (a) By how much does the beam deflect? (b) Does the atomic force microscope have sufficient vertical resolution to detect such a deflection? (c) If the deflected beam formed part of the curved surface of a large virtual sphere, what would be the sphere’s diameter? ANSWER A) C for a rectangular beam is 3. m CLZG —V)A0 m (3X200x10-5m}(1—- 0.22)(2 x10”4 N/m)m AZ _ 02 Eat; (100x109Pafix10‘6m) 19 m B) Yes. Vertical resolution of an AFM is about 0.01 nm. C) the radius of curvature would be: EMtZ ... (100x109 Pafixlo-Gm} R 2 —— — _4 =107m3 so the diameter would be 2 X 107 3 66»va abuzz-limo N/ml 214 m ...
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