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7). mechanical waves (finished)
- 1.
- 2. Mechanical Wave Classification ofwaves (i). Medium ;Mechanical and Electromagnetic waves (ii). Direction of particles of medium ;Transverse and Longitudinal waves (iii). Motion of wave ;Stationary and progressive waves In elastic mediums, Tensile Stress, 𝑺 𝒏 𝑆 𝑛 = 𝐹 𝐴 𝑛𝑜𝑟𝑚𝑎𝑙 𝑃𝑎 ***Similar to pressure ***𝐴 𝑛𝑜𝑟𝑚𝑎𝑙 is the perpendicular area to the force. Tensile Strain, 𝜱 𝒕 Φ 𝑡 = ∆𝐿 𝐿0 Young’s Modulus, 𝒀 𝑌 = 𝑆 𝑛 Φ 𝑡 𝑃𝑎 ***It is vary from one material to another. Same material has the same 𝑌. Shear Stress, 𝑺 𝒕 𝑆𝑡 = 𝐹 𝐴 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑃𝑎 Shear Strain, 𝜱 𝒔 Φ 𝑠 = ∆𝑥 𝐿0 Shear Modulus, 𝑹 𝑅 = 𝑆𝑡 Φ 𝑠 𝑃𝑎 Pressure in fluid 𝐹−𝐹 𝐿0 ∆𝐿 𝐴 𝐹 −𝐹 𝐿0 ∆𝑥 𝐴 𝐹 𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑓𝑖𝑛𝑎𝑙
- 3. Hydrostatic pressure, 𝑷 𝑃= 𝐹 𝐴 𝑃𝑎 ***equivalent to stress Volume Strain, 𝜱 𝒔 Φ 𝑉 = 𝑉𝑓 − 𝑉𝑖 𝑉𝑖 = ∆𝑉 𝑉𝑖 Bulk Modulus, 𝑩 𝐵 = − ∆𝑃 Φ 𝑉 𝑃𝑎 ***𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑖𝑙𝑖𝑡𝑦 = 1 𝐵 Stress-Strain Curve Strain can reflect the force applied 𝐴 → 𝐵; linear variation, elastic behavior ***The gradient = Young’s Modulus 𝐶 → 𝐷; Plastic behavior 𝐵 = 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙 𝑙𝑖𝑚𝑖𝑡 𝐶 = 𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑙𝑖𝑚𝑖𝑡 𝐷 = 𝐹𝑟𝑎𝑐𝑡𝑢𝑟𝑒 𝑝𝑜𝑖𝑛𝑡 𝑆𝑡𝑟𝑒𝑠𝑠 𝑆𝑡𝑟𝑎𝑖𝑛𝐴 𝐵 𝐶 𝐷 Equation of mechanical waves 𝑦 = 𝑓 𝑥 = 𝑥2 For moving wave, 𝑦 = 𝑓 𝑥 ± 𝑠 = 𝑓 𝑥 ± 𝑣𝑡 However, the shape of wave is Sinusoidal or Simple harmonic wave, and not parabola ∴ 𝑦 = 𝑓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘𝑥 For moving wave, 𝑦 = 𝑓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘(𝑥 ± 𝑣𝑡) For wave movingto the right, 𝑦 = 𝑓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘(𝑥 − 𝑣𝑡) For wave movingto the left, 𝑦 = 𝑓 𝑥 = 𝐴 sin 𝑘(𝑥 + 𝑣𝑡) 𝑦 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝐴 = 𝑎𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝑘 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑤𝑎𝑣𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 λ = 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔ℎ𝑡 𝑓 𝑥 = 𝑥2 𝑓 𝑥 = (𝑥 − 𝑣𝑡)2 𝑓 𝑥 = (𝑥 + 𝑣𝑡)2 𝑥 𝑦 𝑦(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛 𝑘(𝑥 ± 𝑣𝑡) 𝑘 = 2𝜋 λ = 𝜔 𝑣
- 4. 𝑦(𝑥, 𝑡) =𝐴 𝑠𝑖𝑛 𝑘(𝑥 ± 𝑣𝑡) Suppose the wave moves to the right. At 𝑡 = 0, 𝑡0 The position of a particle is 𝑥0 ∴ 𝑦(𝑥0, 𝑡0) = 𝐴 𝑠𝑖𝑛 𝑘(𝑥0 − 𝑣𝑡0) The other position of the adjacent particle that is in phase is 𝑥0 + λ ∴ 𝑦 𝑥0 + 𝜆, 𝑡0 = 𝐴 𝑠𝑖𝑛 𝑘 𝑥0 + 𝜆 − 𝑣𝑡0 However, the points which are in phase have the same displacement (𝑦) ∴ 𝑦 𝑥0, 𝑡0 = 𝑦 𝑥0 + λ, 𝑡0 𝐴 𝑠𝑖𝑛 𝑘(𝑥0 − 𝑣𝑡0) = 𝐴 𝑠𝑖𝑛 𝑘 𝑥0 + 𝜆 − 𝑣𝑡0 𝑠𝑖𝑛(𝑘𝑥0 − 𝑘𝑣𝑡0) = 𝑠𝑖𝑛 𝑘𝑥0 + 𝒌𝝀 − 𝑘𝑣𝑡0 ∴ 𝑘𝜆 = 0 𝑜𝑟 𝑘𝜆 = 2𝜋 Since, 𝜆 = 𝑣 𝑓 𝑘𝑣 = 2𝜋𝑓 = 𝜔 Summary 𝑦(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛 𝑘(𝑥 ± 𝑣𝑡) 𝑘 = 2𝜋 λ = 𝜔 𝑣 𝑣 = 𝑓 λ 𝜔 = 2𝑓𝜋 = 2𝜋 𝑇 𝑘 = 𝜔 𝑣 In reality, at 𝑡0, 𝑦 may not equation to 0 Thus, the general equation is, 𝑦(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛(𝑘𝑥 ± 𝜔𝑡 + 𝜙) 𝜙 𝑖𝑠 𝑎 𝑝ℎ𝑎𝑠𝑒 𝑠ℎ𝑖𝑓𝑡 ***All angles are in radian. 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = 𝑦(𝑥, 𝑡) 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 Wave equation 𝜕2 𝑦 𝜕𝑥2 = −𝑘2 𝐴 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡 = −𝑘2 𝑦 𝜕2 𝑦 𝜕𝑡2 = −𝜔2 𝐴 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡 = −𝜔2 𝑦 ∴ 𝜕2 𝑦 𝜕𝑥2 𝜕2 𝑦 𝜕𝑡2 = 𝜔2 𝑘2 = 𝑣2 ∴ 𝜕2 𝑦 𝜕𝑥2 = 1 𝑣2 𝜕2 𝑦 𝜕𝑡2 ***Motion equation which can be arranged in this form is a sinusoidal wave. 𝑦(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛(𝑘𝑥 ± 𝜔𝑡) 𝑢(𝑥0, 𝑡0) = 𝜕𝑦 𝜕𝑡 𝑥=𝑥0,𝑡=𝑡0 𝑎 𝑥0, 𝑡0 = 𝜕𝑢 𝜕𝑡 𝑥=𝑥0,𝑡=𝑡0 = 𝜕2 𝑦 𝜕𝑡2 𝑥=𝑥0,𝑡=𝑡0
- 5. Interference Constructive interference Destructive inference StandingWave Suppose there are 2 identical waves , one moves to the left and the other move to the right, combine together. Wave1; 𝑦1(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛(𝑘𝑥 + 𝜔𝑡) Wave2; 𝑦2(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) ∴ 𝑦 = 𝑦1 + 𝑦1 = 𝐴 𝑠𝑖𝑛(𝑘𝑥 + 𝜔𝑡) + 𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) = 2𝐴 𝑠𝑖𝑛 𝑘𝑥 𝑐𝑜𝑠 𝜔𝑡 Thus, the new amplitude is 2𝐴 𝑠𝑖𝑛 𝑘𝑥. When 𝑠𝑖𝑛 𝑘𝑥 = ±1, Anti-node When 𝑠𝑖𝑛 𝑘𝑥 = 0, Node When 𝑠𝑖𝑛 𝑘𝑥 = ±1, Anti-node 𝐵𝑎𝑠𝑖𝑐 𝑎𝑛𝑔𝑙𝑒 = 𝜋 2 ∴ 𝑘𝑥 = 𝜋 2 , 3𝜋 2 , 5𝜋 2 , … Since 𝑘 = 𝜔 𝑣 = 2𝜋𝑓 𝑓λ = 2𝜋 λ , 2𝜋 λ 𝑥 = 𝜋 2 , 3𝜋 2 , 5𝜋 2 , … 𝑥 = λ 4 , 3λ 4 , 5λ 4 , … When 𝑠𝑖𝑛 𝑘𝑥 = 0, Node 𝐵𝑎𝑠𝑖𝑐 𝑎𝑛𝑔𝑙𝑒 = 0 ∴ 𝑘𝑥 = 0, 𝜋, 3𝜋, … 𝑥 = 0, λ 2 , λ, 3λ 2 , … 𝑦(𝑥, 𝑡) = 2𝐴 𝑠𝑖𝑛 𝑘𝑥 𝑐𝑜𝑠 𝜔𝑡 Anti-node 𝑥 = 𝑛λ 4 ; 𝑛 = 1, 3, 5, …
- 6. Fixed end Freeend Reflected waves changed𝜙 by 𝜋 unchanged 𝜙 At the boundary Node Anti-node Refractive index, 𝑛 low high high low Speed high low low high Reflection of wave 1. Reflection from a fixed/hard boundary 2. Reflection from a free/soft boundary Anti-node 𝑥 = 𝑛λ 4 ; 𝑛 = 1, 3, 5, … Node 𝑥 = 𝑛λ 2 ; 𝑛 = 0, 1, 2, … 𝑎𝑛 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑤𝑎𝑣𝑒 𝑎 𝑟𝑒𝑑𝑙𝑒𝑐𝑡𝑒𝑑 𝑤𝑎𝑣𝑒 𝑎𝑛 𝑖𝑛𝑐𝑖𝑑𝑒𝑛𝑡 𝑤𝑎𝑣𝑒 𝑎 𝑟𝑒𝑑𝑙𝑒𝑐𝑡𝑒𝑑 𝑤𝑎𝑣𝑒 ***a reflected wave is identical to the incident wave except moving in an opposite direction, and a phase shift may change. Thus, the interference of incident waves and reflected waves may result in standing waves. String The fundamental tone λ1 = 2𝑙 𝑓1 = 1 2 𝑣 𝑙 𝑓1 = 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 ***the fundamental frequency is the minimum frequency required to form standing waves. ***the 1st harmonic Harmonic; how many time is the frequency greater than the fundamental frequency, 𝑓1. 𝑁 𝐴 𝐴𝐴 𝑁 𝐴 𝑁𝑁 λ 2 λ 4 𝑁 = 𝑛𝑜𝑑𝑒 𝐴 = 𝑎𝑛𝑡𝑖 − 𝑛𝑜𝑑𝑒 𝑁 𝑁 𝑙
- 7. The first overtone λ2= 𝑙 𝑓2 = 𝑣 𝑙 = 2𝑓1 ***the 2nd harmonic, 𝑓2 𝑓1 = 2. The second overtone λ3 = 2 3 𝑙 𝑓2 = 3 2 𝑣 𝑙 = 3𝑓1 ***the 3rd harmonic, 𝑓2 𝑓1 = 3. ***Natural frequencies are any frequencies required to form standing waves. 𝑙 𝑙 𝑣 = 𝑇 𝜇 Velocity of the string 𝑇 = 𝑠𝑡𝑟𝑖𝑛𝑔 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑁 𝜇 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑘𝑔 𝑚 ***linear density; a measure of mass per unit of length Open ended pipe The fundamental tone λ1 = 2𝑙 𝑓1 = 1 2 𝑣 𝑙 𝑓1 = 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 ***the 1st harmonic The first overtone λ2 = 𝑙 𝑓2 = 𝑣 𝑙 = 2𝑓1 ***the 2nd harmonic, 𝑓2 𝑓1 = 2. 𝑙 𝑙
- 8. The second overtone λ3= 2 3 𝑙 𝑓2 = 3 2 𝑣 𝑙 = 3𝑓1 ***the 3rd harmonic, 𝑓2 𝑓1 = 3. Close ended pipe The fundamental tone λ1 = 4𝑙 𝑓1 = 1 4 𝑣 𝑙 𝑓1 = 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 ***the 1st harmonic The first overtone λ2 = 4 3 𝑙 𝑓2 = 3 4 𝑣 𝑙 = 3𝑓1 ***the 3rd harmonic, 𝑓2 𝑓1 =3. The second overtone λ3 = 4 5 𝑙 𝑓2 = 5 4 𝑣 𝑙 = 5𝑓1 ***the 5th harmonic, 𝑓2 𝑓1 = 5. ***The harmonics of close ended pipe are only odd numbers. Velocity of wave inside the pipe 𝑙 𝑙 𝑙 𝑙 𝑣 = 𝐵 𝜌
- 9. Resonance ; the tendencyof a system to oscillate at a greater amplitude at some frequencies (natural frequencies) than at others. All system has its own frequency. If a force is applied with the same frequency of the system, the damage is largest as it can reach the maximum amplitude. Modulation and Beat The interference of 2 waves with the same amplitude and direction, but different 𝑘 and 𝜔 Wave1; 𝑦1(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛(𝑘1 𝑥 − 𝜔1 𝑡) Wave2; 𝑦2(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛(𝑘2 𝑥 − 𝜔2 𝑡) 𝑦 = 𝑦1 + 𝑦2 𝑦 = 2𝐵 𝑐𝑜𝑠(𝑘 𝑚𝑜𝑑 𝑥 − 𝜔 𝑚𝑜𝑑 𝑡) 𝑠𝑖𝑛(𝑘 𝑎𝑣 𝑥 − 𝜔 𝑎𝑣 𝑡) When, 𝑘 𝑚𝑜𝑑 = 𝑘2−𝑘1 2 ,𝑘 𝑎𝑣 = 𝑘2+𝑘1 2 And, 𝜔 𝑚𝑜𝑑 = 𝜔2−𝜔1 2 ,𝜔 𝑎𝑣 = 𝜔2+𝜔1 2 𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2 𝑓 𝑚𝑜𝑑 = 𝜔 𝑚𝑜𝑑 2𝜋 = 1 2 𝑓2 − 𝑓1 Since, 𝑦 = 2𝐵 𝑐𝑜𝑠(𝑘 𝑚𝑜𝑑 𝑥 − 𝜔 𝑚𝑜𝑑 𝑡) 𝑠𝑖𝑛(𝑘 𝑎𝑣 𝑥 − 𝜔 𝑎𝑣 𝑡) can be expressed in the form of 𝑦 = 𝐴 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡 When 𝐴 = 2𝐵 𝑐𝑜𝑠(𝑘 𝑚𝑜𝑑 𝑥 − 𝜔 𝑚𝑜𝑑 𝑡) ∴ We define, 𝐴 𝑚𝑜𝑑 = 2𝐵 𝑐𝑜𝑠(𝑘 𝑚𝑜𝑑 𝑥 − 𝜔 𝑚𝑜𝑑 𝑡) ***The amplitude of the combined wave is a sinusoidal function also. The average velocity of the combined wave,𝑣 𝑝ℎ𝑎𝑠𝑒 = 𝑣 𝑎𝑣 = 𝜔 𝑎𝑣 𝑘 𝑎𝑣 𝑦𝑡𝑜𝑡𝑎𝑙propagates with 𝑓𝑎 𝑣, 𝑣 𝑎𝑣 The Envelope propagates with 𝑓 𝑚𝑜𝑑, 𝑣𝑔 𝑣𝑔 = 𝜔 𝑚𝑜𝑑 𝑘 𝑚𝑜𝑑 = 𝜔2 − 𝜔1 𝑘2 − 𝑘1 = 𝑑𝜔 𝑑𝑘 𝑣 𝑎𝑣 = 𝜔 𝑎𝑣 𝑘 𝑎𝑣 𝑦𝑡𝑜𝑡𝑎𝑙 𝐸𝑛𝑣𝑒𝑙𝑜𝑝𝑒 (𝐴 𝑚𝑜𝑑) 𝑣𝑔 = 𝜔2 − 𝜔1 𝑘2 − 𝑘1 = 𝑑𝜔 𝑑𝑘
- 10. String 𝑣 = 𝑣𝑒𝑙𝑜𝑠𝑖𝑡𝑦 𝑇= 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑁 𝜇 = 𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑘𝑔 𝑚 The rate of energy transfer, 𝑱 𝒔 𝐸𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐸 + 𝐾𝐸 For Kinetic Energy, 𝐾𝐸 = 1 2 𝑚𝑢2 𝑑𝐾𝐸 = 1 2 𝑑𝑚 𝑢2 Since, 𝑢 = 𝜕𝑦 𝜕𝑡 𝑢 = 𝜕 𝜕𝑡 𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) 𝑢 = 𝐴 𝜕 𝜕𝑡 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) 𝑢 = −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) And 𝜇 = 𝑑𝑚 𝑑𝑥 𝑜𝑟 𝑑𝑚 = 𝜇𝑑𝑥 ∴ 𝑑𝐾𝐸 = 1 2 𝜇𝑑𝑥 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2 𝑑𝐾𝐸 𝑑𝑡 = 1 2 𝜇 𝑑𝑥 𝑑𝑡 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2 𝑑𝐾𝐸 𝑑𝑡 = 1 2 𝜇𝑣𝜔2 𝐴2 𝑐𝑜𝑠2 (𝑘𝑥 − 𝜔𝑡) 𝑑𝐾𝐸 𝑑𝑡 = 1 2 𝜇𝑣𝜔2 𝐴2 𝑐𝑜𝑠2 (𝑘𝑥 − 𝜔𝑡) 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜇𝑣𝜔2 𝐴2 1 𝑇 𝑐𝑜𝑠2 (𝑘𝑥 − 𝜔𝑡) 𝑇 0 𝑑𝑡 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜇𝑣𝜔2 𝐴2 1 2 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 1 4 𝜇𝑣𝜔2 𝐴2 From Equipartition Theorem, 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 𝑑𝑃𝐸 𝑑𝑡 𝑎𝑣 ∴ 𝑑𝐸 𝑑𝑡 𝑎𝑣 = 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 + 𝑑𝑃𝐸 𝑑𝑡 𝑎𝑣 𝑑𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜇𝑣𝜔2 𝐴2 Wave velocity in a fluid 𝐵 = 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑣 = 𝑇 𝜇 𝑑𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜇𝑣𝜔2 𝐴 𝑚𝑎𝑥 2 𝑣 = 𝐵 𝜌
- 11. Longitudinal Waves We haveanother function to represent longitudinal waves, using Pressure 𝑃 𝑥, 𝑡 At (a), the air particles are at equilibrium point with the cross- sectional diameter (of the column of the air ) of 𝐷. When the longitudinal wave is produced in (b). Air particles move with different rate, causing the change in pressure. Since, Bulk Modulus, 𝐵 = − ∆𝑃 Φ 𝑉 And Φ 𝑉 = ∆𝑉 𝑉𝑖 𝐵 = − ∆𝑃 ∆𝑉/𝑉𝑖 (a). (b). 𝑦 𝑦 + ∆𝑦 𝐷 ∆𝑥 𝑝0 𝑝0 + 𝑝 𝐵 = − ∆𝑃 ∆𝑉/𝑉𝑖 𝐵 = − 𝑃𝑓 − 𝑃𝑖 𝑉𝑓 − 𝑉𝑖 /𝑉𝑖 𝐵 = − 𝑝0 + 𝑝 − 𝑝0 ∆𝑥 + ∆𝑦 𝐷 − ∆𝑥𝐷 /∆𝑥𝐷 𝐵 = − 𝑝 ∆𝑦/∆𝑥 𝑝 = −𝐵 ∆𝑦 ∆𝑥 Since, y 𝑥, 𝑡 = 𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) ∴ 𝑝 = −𝐵 𝜕 𝐴 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡 𝜕𝑥 𝑝 = −𝐵𝐴𝑘 𝑐𝑜𝑠 𝑘𝑥 − 𝜔𝑡 𝑝 = −𝑃 𝑐𝑜𝑠 𝑘𝑥 − 𝜔𝑡 We define, 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝑨𝒎𝒑𝒍𝒊𝒕𝒖𝒅𝒆, 𝑃 = 𝐵𝐴𝑘 Since, 𝑣 = 𝐵 𝜌 , 𝐵 = 𝜌𝑣2 ∴ 𝑃 = 𝜌𝑣2 𝐴𝑘 𝑝 = −𝐵 𝜕𝑦 𝜕𝑥 𝑦(𝑥, 𝑡)𝑝 𝑥, 𝑡 −𝐵 𝜕 𝜕𝑥 𝑃 = 𝐵𝐴𝑘 = 𝜌𝑣2 𝐴𝑘
- 12. Since, When y 𝑥,𝑡 = 𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡), 𝑝 = −𝑃 𝑐𝑜𝑠 𝑘𝑥 − 𝜔𝑡 We can express, 𝑝 = 𝑃 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡 − 𝜋 2 Therefore, the shapes of the 2 graphs are the same. However, the 𝑝ℎ𝑎𝑠𝑒 𝑠ℎ𝑖𝑓𝑡, 𝜙 = 𝜋 2 𝑜𝑟 90° The rate of energy transfer or the longitudinal wave 𝐸𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐸 + 𝐾𝐸 For Kinetic Energy, 𝐾𝐸 = 1 2 𝑚𝑢2 𝑑𝐾𝐸 = 1 2 𝑑𝑚 𝑢2 Since, 𝑢 = 𝜕𝑦 𝜕𝑡 𝑢 = 𝜕 𝜕𝑡 𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) 𝑢 = 𝐴 𝜕 𝜕𝑡 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡) 𝑢 = −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) And 𝜌 = 𝑑𝑚 𝑑𝑉 𝑜𝑟 𝑑𝑚 = 𝜌 𝑑𝑣 = 𝜌 𝐷𝑑𝑥 ; 𝐷 = 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 ∴ 𝑑𝐾𝐸 = 1 2 𝜌𝐷𝑑𝑥 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2 𝑑𝐾𝐸 𝑑𝑡 = 1 2 𝜌𝐷 𝑑𝑥 𝑑𝑡 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2 𝑑𝐾𝐸 𝑑𝑡 = 1 2 𝜌𝐷𝑣𝜔2 𝐴2 𝑐𝑜𝑠2(𝑘𝑥 − 𝜔𝑡) 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜌𝐷𝑣𝜔2 𝐴2 1 𝑇 𝑐𝑜𝑠2(𝑘𝑥 − 𝜔𝑡) 𝑇 0 𝑑𝑡 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜌𝐷𝑣𝜔2 𝐴2 1 2 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 1 4 𝜌𝐷𝑣𝜔2 𝐴2 From Equipartition Theorem, 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 = 𝑑𝑃𝐸 𝑑𝑡 𝑎𝑣 ∴ 𝑑𝐸 𝑑𝑡 𝑎𝑣 = 𝑑𝐾𝐸 𝑑𝑡 𝑎𝑣 + 𝑑𝑃𝐸 𝑑𝑡 𝑎𝑣 𝑑𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜌𝐷𝑣𝜔2 𝐴2 𝑑𝐸 𝑑𝑡 𝑎𝑣 = 1 2 𝜌𝐷𝑣𝜔2 𝐴 𝑚𝑎𝑥 2