GSO002 Problem 17
Problem Statement
Energy Distribution in String Vibration via Fourier Series
Problem Statement
In string instruments such as guitars, plucking (picking) a string at a specific location gives the string a particular initial displacement, exciting an infinite number of normal modes. In this problem, we rigorously analyze this phenomenon using the 1D wave equation and Fourier series expansion.
Take the horizontal direction as the x-axis. A flexible, uniform string of linear mass density ρ and tension T is fixed at positions x=0 and x=L. Let the displacement of the string be y(x,t), which obeys the 1D wave equation ρ∂t2∂2y=T∂x2∂2y for small vibrations. Damping due to air resistance or internal friction is neglected.
At time t=0, the point x=3L on the string is pulled perpendicular to the string by distance h and gently released (initial velocity is zero everywhere). The initial shape consists of two line segments: one from x=0 to x=3L, and another from x=3L to x=L, forming a triangle.
The string vibration y(x,t) can be expressed as a superposition of normal modes (Fourier sine series):
y(x,t)=n=1∑∞Ansin(Lnπx)cos(ωnt)where yn(x,t) is the displacement of the n-th harmonic (fundamental at n=1), ωn is its angular frequency, and An is the Fourier coefficient determining the amplitude of each mode.
Define the mechanical energy (sum of kinetic and elastic energy) of the n-th harmonic alone as En.
Under the given constraints, compute the mechanical energy E1 of the fundamental mode (n=1).
Constraints
- L=0.90 [m]
- ρ=5.0×10−3 [kg/m]
- T=160 [N]
- h=0.03 [m]
Input Format
The calculated energy E1 [J] is expressed as:
E1=BπCAwhere A and B are coprime positive integers and C is a positive integer.
Give the value of A+B+C as a positive integer.
Solution
Normal Modes and General Solution
With boundary conditions y(0,t)=y(L,t)=0 and zero initial velocity:
y(x,t)=n=1∑∞Ansin(knx)cos(ωnt)where kn=Lnπ and ωn=LnπρT.
Fourier Coefficients
The initial displacement is:
f(x)={L3hx2L3h(L−x)(0≤x≤3L)(3L<x≤L)The Fourier coefficient:
An=L2∫0Lf(x)sin(knx)dxAfter integration by parts (the cos terms cancel):
An=n2π29hsin(3nπ)Mechanical Energy of Each Mode
Since the system is conservative, En equals the maximum potential energy at t=0:
En=21T∫0L(Ankncos(knx))2dx=41TLkn2An2=4Ln2π2TAn2Substituting An:
En=4n2π2L81Th2sin2(3nπ)Numerical Calculation for n=1
E1=4π2L81Th2sin2(3π)=4π2L81Th2⋅43=16π2L243Th2Substituting T=160, h2=9×10−4, L=0.90:
E1=16×π2×0.90243×160×9×10−4=14.4π2243×0.144=100π2243 [J]Comparing with E1=BπCA: A=243, B=100, C=2 (gcd(243,100)=1).
A+B+C=243+100+2=345Final Answer: 345