Transcript Slide 1
Standing Waves • • • • • • • • Review Harmonic Oscillator Review Traveling Waves Formation of Standing Waves Standing Waves and Resonance Resonance Variables String instruments Examples Standing Waves and Electron Orbitals Brazilian guitarist Badi Assad Summary - Simple Harmonic Oscillator • Energy 1 2 1 2 1 2 1 2 𝐸 = 𝑘𝑥 2 + 𝑚𝑣 2 = 𝑘𝐴2 = 𝑘𝑣𝑚𝑎𝑥 2 • Motion 𝑥 = 𝐴𝑠𝑖𝑛 𝜔𝑡 𝑜𝑟 𝑥 = 𝐴𝑐𝑜𝑠 𝜔𝑡 • Harmonic frequency 𝜔= 𝑘 𝑚 • Frequency and period 𝜔 = 2𝜋𝑓 𝜔 = 2𝜋 𝑇 Summary – Traveling Waves • Behaves as coupled harmonic oscillator – Motion of individual oscillators – Motion of disturbance between oscillators • Wave equation for string – F = ma on individual segment – Sinusoidal traveling solutions (also pulse) – Velocity 𝑣 = 𝑇 𝜇 • Properties of sinusoidal waves – – – – Amplitude Wavelength Frequency/Period Velocity 𝑣 = 𝑓𝜆 • Other topics – Transverse vs. longitudinal – Energy spreading for spherical wave Traveling wave animation http://www.animations.physics.unsw.edu.au/waves-sound/travelling-wavesII Standing wave animation • Animation http://faraday.physics.utoronto.ca/IYearLab/Intros/StandingWaves/Flash/reflect.html • Note Peaks oscillate in place – Nodes – Antinodes • Only certain wavelengths “fit” – Basis for stringed instruments (guitar/violin/piano) – Electromagnetic resonances – Quantum mechanics Formation of Standing Wave • Add incident wave traveling to right 𝑌𝑖𝑛𝑐 = 𝐴𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) • With reflected wave traveling to left 𝑌𝑟𝑒𝑓 = 𝐴𝑠𝑖𝑛(𝑘𝑥 + 𝜔𝑡) • And use trig identify: sin 𝑎 ± 𝑏 = sin 𝑎 cos(𝑏) ± cos 𝑎 sin(𝑏) 𝑌𝑡𝑜𝑡 = 𝑌𝑖𝑛𝑐 + 𝑌𝑟𝑒𝑓 = 2𝐴𝑠𝑖𝑛 𝑘𝑥 cos(𝜔𝑡) • Produces standing wave – oscillates in place Another Standing wave animation • Can turn on/off reflection http://phet.colorado.edu/sims/wave-on-a-string/wave-on-a-string_en.html • Settings – – – – Oscillate No end / fixed end Small, but some damping Turn amplitude down to 1 • Note: almost gets out of control! Standing wave with both fixed ends • Both ends (approximately) fixed http://faraday.physics.utoronto.ca/IYearLab/Intros/StandingWaves/Flash/sta2fix.html • Only certain wavelengths “fit” Standing waves and resonance • If string fixed both ends, only certain wavelengths “fit” 𝑳= 𝝀 𝟐 𝑳=𝝀 𝑳= 𝟑 𝑳 𝟐 Standing waves and resonance • String anchored between 2 points • Allowed opening widths 𝜆 – 𝐿=2 – 𝐿=𝜆 – 𝐿= 3𝜆 2 • In general – 𝐿= 𝑛𝜆 2 𝑛 = 1,2,3 … • Allowed wavelengths – 𝜆𝑛 = 2𝐿 𝑛 𝑛 = 1,2,3…. • Allowed frequencies 𝑣 – 𝑓𝑛 = 𝜆 = 𝑛 𝑛𝑣 2𝐿 𝑛 = 1,2,3 … . Stringed Instruments • 2 key equations – 𝑓𝑛 = – 𝑣= 𝑣 𝜆𝑛 = 𝑛𝑣 2𝐿 𝑛 = 1,2,3 … . 𝑇 𝜇 • Factors effecting frequency – String length (guitar frets) – Tension (violin tuning) – Mass/length (guitar vs. bass) Example 11-14 • Wave velocity needed for length, frequency, and n 𝑣 1𝑣 𝑓1 = 𝜆 = 2𝐿 1 𝑣 = 2𝐿𝑓1 = 2.2 𝑚 131 𝑠 = 288 𝑚 𝑠 • Mass/length 𝜇= .009 𝑘𝑔 1.1 𝑚 = 0.0082 𝑘𝑔 𝑚 • Tension needed for velocity and mass/length 𝑣= • 𝑇 𝜇 Harmonics 𝑓𝑛 = 𝑛𝑓1 𝑇 = 𝑣 2 𝜇 = 288 𝑚 𝑠 2 .0082 𝑘𝑔 𝑚 = 679 𝑁 = 131, 261, 393, 524 𝐻𝑧 Problem 53 • Assume same v and n for both frequencies. • Write original frequency 𝑓𝑛 = • Write shortened frequency 𝑓𝑛 ′ = • 𝑛𝑣 2𝐿 𝑛𝑣 2 2 3𝐿 Tale the ratio 𝑛𝑣 2 𝑓′ 2 3 𝐿 3 = 𝑛𝑣 = 𝑓 2 2𝐿 𝑓′ = 3 294 𝐻𝑧 = 441 𝐻𝑧 2 Problem 56 • Write harmonic (n) as function of fundamental 𝑓𝑛 = • 𝑛𝑣 2𝐿 Write harmonic (n+1) as function of fundamental 𝑓𝑛+1 = • = 𝑛𝑓1 (𝑛+1)𝑣 2𝐿 = (𝑛 + 1)𝑓1 Subtract the difference 𝑓𝑛+1 − 𝑓𝑛 = 𝑛 + 1 𝑓1 − 𝑛𝑓1 = 𝑓1 𝑓𝑛+1 − 𝑓𝑛 = 350 𝐻𝑧 − 280 𝐻𝑧 = 70 𝐻𝑧 = 𝑓1 Problem 58 • Combining frequency and velocity equations 𝑓𝑛 = 𝑛𝑣 2𝐿 = 𝑇 𝜇 𝑛 2𝐿 • After tuning 𝑓𝑛 ′ = 𝑛𝑣 2𝐿 = 𝑛 𝑇′ 𝜇 2𝐿 • Ratio 𝑓′ 𝑓 = 𝑇′ 𝑇 𝑇′ 𝑇 = 𝑓′2 𝑓2 = 2002 2052 = 95.2 % Decrease 4.8% Problem 59 • Find velocity needed for number of antinodes • Find tension needed for those velocities Problem 59 (2) • Hold frequency constant and vary velocity with n • Allowed wavelengths 𝜆𝑛 = • 2𝐿 𝑛 Normally we do allowed frequencies in terms of allowed wavelengths 𝑓𝑛 = • 𝑛 = 1,2,3…. 𝑣 𝜆𝑛 = 𝑛𝑣 2𝐿 Now we do allowed velocities in terms of allowed wavelengths, with frequency constant 𝑣𝑛 = 𝑓𝜆𝑛 = 𝑓 • The we do allowed tensions, assuming frequency constant 𝑣= • 2𝐿 𝑛 𝑇 𝜇 Masses are thus 𝑇𝑛 = 𝜇𝑣𝑛 2 = 4𝜇𝑓2 𝐿2 𝑛2 𝑚𝑛 = 𝑇𝑛 𝑔 = 4𝜇𝑓2 𝐿2 𝑔 𝑛2 Problem 59 (3) • For 1 loop 𝑚𝑛 = 4𝜇𝑓2 𝐿2 𝑔 𝑛2 = 4 3.9∙10−4 𝑘𝑔 𝑚 60 𝑠 2 1.5 𝑚 2 9.8 𝑚 𝑠 2 12 = 1.29 𝑘𝑔 • For 2 loops 𝑚𝑛 = 4𝜇𝑓2 𝐿2 𝑔 𝑛2 = 4 3.9∙10−4 𝑘𝑔 𝑚 60 𝑠 2 1.5 𝑚 2 9.8 𝑚 𝑠 2 22 = 0.32 𝑘𝑔 • For 5 loops 𝑚𝑛 = 4𝜇𝑓2 𝐿2 𝑔 𝑛2 = 4 3.9∙10−4 𝑘𝑔 𝑚 60 𝑠 2 1.5 𝑚 2 9.8 𝑚 𝑠 2 52 = 0.052 𝑘𝑔 Standing waves and electron orbitals http://www.upscale.utoronto.ca/PVB/Harrison/Flash/QuantumMechanics/CircularStandWaves/ CircularStandWaves.html (sorry the downloadable .swf doesn’t seem to work)