Wave Motion and String Waves Concept Page - 5

Example

Derive and find kinetic energy per unit length at a given point in a travelling string wave

A uniform string of length / is fixed at both ends such that tension T is produced in it. The string is excited to vibrate with maximum displacement amplitude ao$a_o$. The maximum kinetic energy of the string for its fundamental tone is given as a20π2Txl${\displaystyle {\displaystyle \frac{a_0^2{\pi }^{2}T}{xl}}}$. Find x$x$.

Solution:
Since tension on the string is T$T$ so the velocity of the wave should be v=Tm−−−√$v=\sqrt{{\displaystyle \frac{T}{m}}}$, where m$m$ is mass per unit length.
Since frequency is n=vλ$n={\displaystyle \frac{v}{\lambda }}$  there for the frequency of the fundamental tone should be n1=12lTm−−−√$n_1={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}$
Now consider an elemental length of a string at a distance x$x$ of size dx$dx$.
So its mass is: dm=mdx$dm=mdx$
Its oscillation energy is given by 12(mdx)a2(2πn1)2${\displaystyle \frac{1}{2}}(mdx)a^2({2\pi n_1)}^2$ =a20π2T2l2sin2(2πx2l)dx$={\displaystyle \frac{a_0^2{\pi }^{2}T}{2l^2}}{\sin }^2({\displaystyle \frac{2\pi x}{2l}})dx$      
[n1=12lTm−−−√$n_1={\displaystyle \frac{1}{2l}}\sqrt{{\displaystyle \frac{T}{m}}}$ ]
Integrating this we get, total energy as a20π2T4l${\displaystyle \frac{a_0^2{\pi }^{2}T}{4l}}$
This gives us x=4$x=4$

Definition

Potential energy in a travelling string wave

Potential energy/meter=∫12T(∂y∂x)2dx=∫12TA2k2cos2(kx−ωt)dx$\int {\displaystyle \frac{1}{2}}T{\left({\displaystyle \frac{\mathrm{\partial }y}{\mathrm{\partial }x}}\right)}^{2}dx=\int {\displaystyle \frac{1}{2}}T{A}^2{k}^2{\cos }^2(kx-\omega t)dx$
The average potential energy per meter of string is=14TA2k2=14μω2A2${\displaystyle \frac{1}{4}}T{A}^2{k}^2={\displaystyle \frac{1}{4}}\mu {\omega }^2{A}^2$ since ω=vk$\omega =vk$ and v2=Tμ$v^2={\displaystyle \frac{T}{\mu }}$

Example

Derive and find total kinetic energy in one wavelength in a travelling string wave

Example:
A uniform string of length l$l$ is fixed at both ends such that tension T$T$ is produced in it. The string is excited to vibrate with maximum displacement amplitude ao$a_o$. The kinetic energy of the string for its first overtone is given as a20π2Txl${\displaystyle {\displaystyle \frac{a_0^2{\pi }^{2}T}{xl}}}$. Find x$x$

Solution:
Displacement equation of a stationary wave          y=2Asin(knx)sin(wnt)$y=2Asin(k_{n}x)sin(w_{n}t)$
First overtone means   n=2$n=2$
Kinetic energy of a small element      dE=12(μdx)(dydt)2$dE={\displaystyle \frac{1}{2}}(\mu dx)({\displaystyle \frac{dy}{dt}}{)}^2$
⟹$⟹$   dE=12(μdx)(2Asin(knx))2×w2ncos(wnt)$dE={\displaystyle \frac{1}{2}}(\mu dx)(2Asin(k_{n}x){)}^2\times w_n^{2}cos(w_{n}t)$
∴$\therefore$   dE=2A2w2nμsin2(knx)dx$dE=2A^2{w}_n^2\mu si{n}^2(k_{n}x)dx$           .........(1)
Given :        A=ao$A=a_o$  
 Also first overtone frequency          ν2=22lTμ−−√${\nu }_2={\displaystyle \frac{2}{2l}}\sqrt{{\displaystyle \frac{T}{\mu }}}$      
   ⟹$⟹$   ν22=Tμl2${\nu }_2^2={\displaystyle \frac{T}{\mu l^2}}$
As      wn=2πν2$w_n=2\pi {\nu }_2$

∴$\therefore$   w2n=4π2ν22=4π2×Tμl2$w_n^2=4{\pi }^2{\nu }_2^2=4{\pi }^2\times {\displaystyle \frac{T}{\mu l^2}}$

Also    kn=nπx2l=2πx2l=πxl$k_n={\displaystyle \frac{n\pi x}{2l}}={\displaystyle \frac{2\pi x}{2l}}={\displaystyle \frac{\pi x}{l}}$

Putting these values in  (1) we get,      
  dE=2a2oμ×(4π2×Tμl2)sin2(knx)dx$dE=2a_o^2\mu \times (4{\pi }^2\times {\displaystyle \frac{T}{\mu l^2}})si{n}^2(k_{n}x)dx$

⟹$⟹$   dE=8a2oπ2Tl2sin2(πxl)dx$dE={\displaystyle \frac{8a_o^2{\pi }^{2}T}{l^2}}si{n}^2({\displaystyle \frac{\pi x}{l}})dx$

Total kinetic energy        E=8a2oπ2Tl2∫sin2(πxl)dx$E={\displaystyle \frac{8a_o^2{\pi }^{2}T}{l^2}}\int si{n}^2({\displaystyle \frac{\pi x}{l}})dx$

   E=8a2oπ2Tl2∫l01−cos(2πx/l)2dx$E={\displaystyle \frac{8a_o^2{\pi }^{2}T}{l^2}}{\int }_0^l\frac{1-cos(2\pi x/l)}{2}dx$   
E=8a2oπ2Tl2×12×[x∣∣∣l0−sin(2πx/l)(2πx/2)∣∣∣l0]$E={\displaystyle \frac{8a_o^2{\pi }^{2}T}{l^2}}\times {\displaystyle \frac{1}{2}}\times [x{|}_0^l-{\displaystyle \frac{sin(2\pi x/l)}{(2\pi x/2)}}{|}_0^l]$   
E=8a2oπ2Tl2×12×[l−0]$E={\displaystyle \frac{8a_o^2{\pi }^{2}T}{l^2}}\times {\displaystyle \frac{1}{2}}\times [l-0]$

  ⟹$⟹$   E=4a2oπ2Tl$E={\displaystyle \frac{4a_o^2{\pi }^{2}T}{l}}$   ∴$\therefore$  x=4$x=4$

Example

Power delivered at a point and total power delivered during one time period in a travelling string wave

Example: A stretched rope having linear mass density 5×10−2 kgm−1$5\times {10}^{-2}kg{m}^{-1}$ is under a tension of 80N$80N$. Find the power that has to be supplied to the rope to generate harmonic waves at a frequency of 60 Hz$60Hz$ and an amplitude of 6 cm$6cm$ ?

Solution:
Velocity v=Tμ−−√=805×10−2−−−−−−−−√=40 ms−1$v=\sqrt{{\displaystyle \frac{T}{\mu }}}=\sqrt{{\displaystyle \frac{80}{5\times {10}^{-2}}}}=40m{s}^{-1}$
Power P=12μω2A2v$P={\displaystyle \frac{1}{2}}\mu {\omega }^2{A}^{2}v$
                =2π2μvA2f2$=2{\pi }^2\mu v{A}^2{f}^2$
                =2π2×5×10−2×40×36×10−4×(60)2$=2{\pi }^2\times 5\times {10}^{-2}\times 40\times 36\times {10}^{-4}\times (60{)}^2$
                ≈512 W$\approx 512W$

Definition

Define intensity of a travelling wave and use its relation with amplitude, frequency or distance from source

The Intensity of waves is defined as the power delivered per unit area. Intensity of wave is proportional to the square of amplitude of the wave. 
Intensity = PowerArea${\displaystyle \frac{Power}{Area}}$Intensity = Po22ρv${\displaystyle \frac{{P_o}^2}{2\rho v}}$ ; where Po=ρvAω$P_o=\rho vA\omega$
ρv=z=$\rho v=z=$ acoustic impedance; A=$A=$ amplitude; ω=$\omega =$ angular frquency;ρ=$\rho =$ density of material; v=$v=$ wave speed

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