VII Mechanical Wave

Mechanical Wave Classification of waves (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, 𝑺𝒏

𝑆𝑛 =𝐹

𝐴𝑛𝑜𝑟𝑚𝑎𝑙𝑃𝑎

***Similar to pressure ***𝐴𝑛𝑜𝑟𝑚𝑎𝑙 is the perpendicular area to the force. Tensile Strain, 𝜱𝒕

Φ𝑡 =∆𝐿

𝐿0

Young’s Modulus, 𝒀

𝑌 =𝑆𝑛

Φ𝑡𝑃𝑎

***It is vary from one material to another. Same material has the same 𝑌.

Shear Stress, 𝑺𝒕

𝑆𝑡 =𝐹

𝐴𝑡𝑎𝑛𝑔𝑒𝑛𝑡𝑃𝑎

Shear Strain, 𝜱𝒔

Φ𝑠 =∆𝑥

𝐿0

Shear Modulus, 𝑹

𝑅 =𝑆𝑡

Φ𝑠𝑃𝑎

Pressure in fluid

𝐹 −𝐹

𝐿0 ∆𝐿

𝐴

𝐹

−𝐹

𝐿0

∆𝑥 𝐴

𝐹

𝑉𝑖𝑛𝑖𝑡𝑖𝑎𝑙

𝑉𝑓𝑖𝑛𝑎𝑙

Hydrostatic pressure, 𝑷

𝑃 =𝐹

𝐴𝑃𝑎

***equivalent to stress Volume Strain, 𝜱𝒔

Φ𝑉 =𝑉𝑓 − 𝑉𝑖

𝑉𝑖=

∆𝑉

𝑉𝑖

Bulk Modulus, 𝑩

𝐵 = −∆𝑃

Φ𝑉𝑃𝑎

***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

𝑆𝑡𝑟𝑎𝑖𝑛 𝐴

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 moving to the right,

𝑦 = 𝑓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘(𝑥 − 𝑣𝑡) For wave moving to 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

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

𝜕2𝑦

𝜕𝑥2= −𝑘2 𝐴 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡 = −𝑘2𝑦

𝜕2𝑦

𝜕𝑡2= −𝜔2 𝐴 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡 = −𝜔2𝑦

∴

𝜕2𝑦𝜕𝑥2

𝜕2𝑦𝜕𝑡2

=𝜔2

𝑘2= 𝑣2

∴𝜕2𝑦

𝜕𝑥2=

1

𝑣2

𝜕2𝑦

𝜕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

=𝜕2𝑦

𝜕𝑡2 𝑥=𝑥0,𝑡=𝑡0

Interference Constructive interference Destructive inference

Standing Wave 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

λ𝑥 =

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,…

Fixed end Free end

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

𝐴 𝐴

𝑁

𝐴 𝑁 𝑁

λ

2

λ

4

𝑁 = 𝑛𝑜𝑑𝑒

𝐴 = 𝑎𝑛𝑡𝑖 − 𝑛𝑜𝑑𝑒

𝑁

𝑁

𝑙

The first overtone

λ2 = 𝑙

𝑓2 =𝑣

𝑙= 2𝑓1

***the 2nd harmonic, 𝑓2

𝑓1= 2.

The second overtone

λ3 =2

3𝑙

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

The second overtone

λ3 =2

3𝑙

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

3𝑙

𝑓2 =3

4

𝑣

𝑙= 3𝑓1

***the 3rd harmonic, 𝑓2

𝑓1=3.

The second overtone

λ3 =4

5𝑙

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

Resonance ; the tendency of 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(𝑘1𝑥 − 𝜔1𝑡) Wave2;

𝑦2(𝑥, 𝑡) = 𝐴 𝑠𝑖𝑛(𝑘2𝑥 − 𝜔2𝑡)

𝑦 = 𝑦1 + 𝑦2 𝑦= 2𝐵 𝑐𝑜𝑠(𝑘𝑚𝑜𝑑𝑥 − 𝜔𝑚𝑜𝑑𝑡) 𝑠𝑖𝑛(𝑘𝑎𝑣𝑥 − 𝜔𝑎𝑣𝑡)

When, 𝑘𝑚𝑜𝑑 =𝑘2−𝑘1

2,𝑘𝑎𝑣 =

𝑘2+𝑘1

2

And, 𝜔𝑚𝑜𝑑 =𝜔2−𝜔1

2,𝜔𝑎𝑣 =

𝜔2+𝜔1

2

𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2

𝑓𝑚𝑜𝑑 =𝜔𝑚𝑜𝑑

2𝜋=

1

2𝑓2 − 𝑓1

Since, 𝑦= 2𝐵 𝑐𝑜𝑠(𝑘𝑚𝑜𝑑𝑥 − 𝜔𝑚𝑜𝑑𝑡) 𝑠𝑖𝑛(𝑘𝑎𝑣𝑥 − 𝜔𝑎𝑣𝑡)

can be expressed in the form of 𝑦 = 𝐴 𝑠𝑖𝑛 𝑘𝑥 − 𝜔𝑡

When 𝐴 = 2𝐵 𝑐𝑜𝑠(𝑘𝑚𝑜𝑑𝑥 − 𝜔𝑚𝑜𝑑𝑡)

∴ We define, 𝐴𝑚𝑜𝑑 = 2𝐵 𝑐𝑜𝑠(𝑘𝑚𝑜𝑑𝑥 − 𝜔𝑚𝑜𝑑𝑡)

***The amplitude of the combined wave is a sinusoidal function also. The average velocity of the combined

wave,𝑣𝑝ℎ𝑎𝑠𝑒 = 𝑣𝑎𝑣 =𝜔𝑎𝑣

𝑘𝑎𝑣

𝑦𝑡𝑜𝑡𝑎𝑙propagates with 𝑓𝑎𝑣, 𝑣𝑎𝑣 The Envelope propagates with 𝑓𝑚𝑜𝑑 , 𝑣𝑔

𝑣𝑔 =𝜔𝑚𝑜𝑑

𝑘𝑚𝑜𝑑=

𝜔2 − 𝜔1

𝑘2 − 𝑘1=

𝑑𝜔

𝑑𝑘

𝑣𝑎𝑣 =𝜔𝑎𝑣

𝑘𝑎𝑣

𝑦𝑡𝑜𝑡𝑎𝑙

𝐸𝑛𝑣𝑒𝑙𝑜𝑝𝑒 (𝐴𝑚𝑜𝑑)

𝑣𝑔 =𝜔2 − 𝜔1

𝑘2 − 𝑘1=

𝑑𝜔

𝑑𝑘

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

2𝑚𝑢2

𝑑𝐾𝐸 =1

2𝑑𝑚 𝑢2

Since, 𝑢 =𝜕𝑦

𝜕𝑡

𝑢 =𝜕

𝜕𝑡𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡)

𝑢 = 𝐴𝜕

𝜕𝑡𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡)

𝑢 = −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) And

𝜇 =𝑑𝑚

𝑑𝑥 𝑜𝑟 𝑑𝑚 = 𝜇𝑑𝑥

∴ 𝑑𝐾𝐸 =1

2𝜇𝑑𝑥 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2

𝑑𝐾𝐸

𝑑𝑡=

1

2𝜇

𝑑𝑥

𝑑𝑡 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2

𝑑𝐾𝐸

𝑑𝑡=

1

2𝜇𝑣𝜔2𝐴2 𝑐𝑜𝑠2(𝑘𝑥 − 𝜔𝑡)

𝑑𝐾𝐸

𝑑𝑡=

1

2𝜇𝑣𝜔2𝐴2 𝑐𝑜𝑠2(𝑘𝑥 − 𝜔𝑡)

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=1

2𝜇𝑣𝜔2𝐴2

1

𝑇 𝑐𝑜𝑠2(𝑘𝑥 − 𝜔𝑡)

𝑇

0

𝑑𝑡

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=1

2𝜇𝑣𝜔2𝐴2

1

2

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=1

4𝜇𝑣𝜔2𝐴2

From Equipartition Theorem,

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=𝑑𝑃𝐸

𝑑𝑡𝑎𝑣

∴𝑑𝐸

𝑑𝑡𝑎𝑣

=𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

+𝑑𝑃𝐸

𝑑𝑡𝑎𝑣

𝑑𝐸

𝑑𝑡𝑎𝑣

=1

2𝜇𝑣𝜔2𝐴2

Wave velocity in a fluid

𝐵 = 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝜌 = 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑

𝑣 =𝑇

𝜇

𝑑𝐸

𝑑𝑡𝑎𝑣

=1

2𝜇𝑣𝜔2𝐴𝑚𝑎𝑥

2

𝑣 =𝐵

𝜌

Longitudinal Waves We have another 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

Φ𝑉

And Φ𝑉 =∆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 = 𝜌𝑣2𝐴𝑘

𝑝 = −𝐵𝜕𝑦

𝜕𝑥

𝑦(𝑥, 𝑡) 𝑝 𝑥, 𝑡 −𝐵

𝜕

𝜕𝑥

𝑃 = 𝐵𝐴𝑘 = 𝜌𝑣2𝐴𝑘

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

2𝑚𝑢2

𝑑𝐾𝐸 =1

2𝑑𝑚 𝑢2

Since, 𝑢 =𝜕𝑦

𝜕𝑡

𝑢 =𝜕

𝜕𝑡𝐴 𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡)

𝑢 = 𝐴𝜕

𝜕𝑡𝑠𝑖𝑛(𝑘𝑥 − 𝜔𝑡)

𝑢 = −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) And

𝜌 =𝑑𝑚

𝑑𝑉 𝑜𝑟 𝑑𝑚 = 𝜌 𝑑𝑣 = 𝜌 𝐷𝑑𝑥

; 𝐷 = 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎

∴ 𝑑𝐾𝐸 =1

2𝜌𝐷𝑑𝑥 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2

𝑑𝐾𝐸

𝑑𝑡=

1

2𝜌𝐷

𝑑𝑥

𝑑𝑡 −𝜔𝐴 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) 2

𝑑𝐾𝐸

𝑑𝑡=

1

2𝜌𝐷𝑣𝜔2𝐴2 𝑐𝑜𝑠2(𝑘𝑥 − 𝜔𝑡)

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=1

2𝜌𝐷𝑣𝜔2𝐴2

1

𝑇 𝑐𝑜𝑠2(𝑘𝑥 − 𝜔𝑡)

𝑇

0

𝑑𝑡

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=1

2𝜌𝐷𝑣𝜔2𝐴2

1

2

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=1

4𝜌𝐷𝑣𝜔2𝐴2

From Equipartition Theorem,

𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

=𝑑𝑃𝐸

𝑑𝑡𝑎𝑣

∴𝑑𝐸

𝑑𝑡𝑎𝑣

=𝑑𝐾𝐸

𝑑𝑡𝑎𝑣

+𝑑𝑃𝐸

𝑑𝑡𝑎𝑣

𝑑𝐸

𝑑𝑡𝑎𝑣

=1

2𝜌𝐷𝑣𝜔2𝐴2

𝑑𝐸

𝑑𝑡𝑎𝑣

=1

2𝜌𝐷𝑣𝜔2𝐴𝑚𝑎𝑥

2