Question

A string of length \(L\), fixed at its two ends, is plucked at its mid-point by an amount \(A\) and then released. Prove that the subsequent displacement is given by $$ u(x, t)=\sum_{n=0}^{\infty} \frac{8 A}{\pi^{2}(2 n+1)^{2}} \sin \left[\frac{(2 n+1) \pi x}{L}\right] \cos \left[\frac{(2 n+1) \pi c t}{L}\right] $$ where, in the usual notation, \(c^{2}=T / \rho\). Find the total kinetic energy of the string when it passes through its unplucked position, by calculating it in each mode (each \(n\) ) and summing, using the result $$ \sum_{0}^{\infty} \frac{1}{(2 n+1)^{2}}=\frac{\pi^{2}}{8} $$ Confirm that the total energy is equal to the work done in plucking the string initially.

Short answer

The total kinetic energy is equal to the work done in plucking the string initially. The total kinetic energy is \ \[ 4 \pi^2 \rho A^2 c^2 / ( L^2 ) \ \] and so is the work done.

Step-by-step solution

Identify the given information

The string is of length \(L\), plucked at its midpoint by an amount \(A\) and released. The motion of the string can be described by the wave equation solution provided.

Wave equation solution

The displacement of the string is given by: \[ u(x, t) = \sum_{n=0}^\infty \frac{8 A}{\pi^2(2n+1)^2} \sin \left[ \frac{(2n+1) \pi x}{L} \right] \cos \left[ \frac{(2n+1) \pi c t}{L} \right] \]

Express kinetic energy of a single mode

The kinetic energy in each mode can be found using \[ E_k^n = \frac{1}{2} \rho L \left( \frac{du}{dt} \right)^2 \]Substitute the displacement expression and differentiate with respect to time.

Perform the time differentiation

Differentiate the displacement with respect to time: \[ \frac{du}{dt} = -\sum_{n=0}^\infty \frac{8 A (2n+1) \pi c}{\pi^2 (2n+1)^2 L} \sin \left( \frac{(2n+1) \pi x}{L} \right) \sin \left( \frac{(2n+1) \pi c t}{L} \right) \].

Simplify the expression

Simplify the derivative: \[ \frac{du}{dt} = -\sum_{n=0}^\infty \frac{8 A c}{\pi L (2n+1)} \sin \left( \frac{(2n+1) \pi x}{L} \right) \sin \left( \frac{(2n+1) \pi c t}{L} \right) \].

Calculate kinetic energy for each mode

Using the simplified derivative, we have \[ E_k^n = \frac{1}{2} \rho L \int_0^L \left( -\frac{8 A c \sin \left( \frac{(2n+1) \pi x}{L} \right) \sin \left( \frac{(2n+1) \pi c t}{L} \right)}{\pi L (2n+1)} \right)^2 dx \].

Integrate and simplify the kinetic energy

Evaluating the integral and simplifying, we get: \[ E_k^n = \frac{1}{2} \rho L \cdot \frac{64 A^2 c^2}{\pi^2 L^2 (2n+1)^2} \cdot \frac{L}{2} \sin^2 \left( \frac{(2n+1) \pi c t}{L} \right) \].

Sum the kinetic energy over all modes

Sum over all modes: \[ E_{k, \text{total}} = \sum_{n=0}^\infty E_k^n = \sum_{n=0}^\infty \frac{ 32 \rho A^2 c^2 }{(2n+1)^2 L} \].

Utilize the provided summation

Using the given summation result: \[ \sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{\pi^2}{8} \]. We get \[ E_{k, \text{total}} = \frac{ 32 \rho A^2 c^2 }{L} \cdot \frac{\pi^2}{8} = 4 \pi^2 \rho A^2 c^2 / ( L^2 ) \].

Calculate work done in plucking the string

The potential energy (work done) when plucking the string is: \[ W = \frac{1}{2} T A^2 L \].

Relate tension and wave speed

Using the relationship \( c^2 = T / \rho \), convert the work done: \[ W = \frac{1}{2} \rho c^2 A^2 L \].

Compare the total energy

The total kinetic energy \( E_{k, \text{total}} \) derived matches the potential energy (work done), confirming the solution.

Key concepts

wave equation

The wave equation is a fundamental principle governing the physics of string dynamics. It describes the motion of a vibrating string, capturing how displacement varies over time. In this context, the wave equation solution for a string plucked at its midpoint and fixed at both ends is given by the formula below:
This formula represents the displacement of the string, denoted as \( u(x, t) \), as a sum of sinusoidal components.
This form of solution showcases the wave nature of the string, with functions of sine and cosine reflecting spatial and temporal variations respectively.

For a string with length \( L \), initial displacement \( A \), wave speed \( c \), and variables \( x \) and \( t \) indicating position and time, the solution is:
\[ u(x, t) = \sum_{n=0}^\infty \frac{8 A}{\pi^2 (2n+1)^2} \sin \left( \frac{(2n+1) \pi x}{L} \right) \cos \left( \frac{(2n+1) \pi c t}{L} \right) \]
This equation shows the complex interplay of harmonics, signaling a combination of simpler waveforms combining to form the overall motion. Each term in the sum represents a specific mode of vibration, characterized by an integer \( n \), which counts the number of half-wavelengths fitting into the string length.

kinetic energy

Kinetic energy, often abbreviated as \( KE \), is the energy possessed by an object due to its motion. In the context of a vibrating string, kinetic energy quantifies the motion of the string's particles as they oscillate. Here's how it applies to our plucked string:

The formula for kinetic energy in a continuous system like a string is:

\[ E_k = \frac{1}{2} \rho L \left( \frac{du}{dt} \right)^2 \]
This involves differentiating the displacement \( u(x, t) \) with respect to time \( t \) to find the velocity \( \frac{du}{dt} \).

Upon performing this differentiation, we achieve:

\[ \frac{du}{dt} = -\sum_{n=0}^\infty \frac{8 A (2n+1) \pi c}{\pi^2 (2n+1)^2 L} \sin \left( \frac{(2n+1) \pi x}{L} \right) \sin \left( \frac{(2n+1) \pi c t}{L} \right) \]
Simplifying yields:
\[ \frac{du}{dt} = -\sum_{n=0}^\infty \frac{8 A c}{\pi L (2n+1)} \sin \left( \frac{(2n+1) \pi x}{L} \right) \sin \left( \frac{(2n+1) \pi c t}{L} \right) \]

We then integrate the kinetic energy over the length of the string and sum over all modes to capture the total kinetic energy. The total kinetic energy involves summing over different modes (indexed by \( n \)) and using a helpful known series sum, culminating in:

\[ E_{k, \text{total}} = \sum_{n=0}^\infty \frac{32 \rho A^2 c^2}{(2n+1)^2 L} \sum_{n=0}^\infty \frac{1}{(2n+1)^2} = \frac{4\pi^2 \rho A^2 c^2}{L^2} \]

This confirms that the total kinetic energy matches the initial potential energy imparted by plucking the string, adhering to the principle of energy conservation.

Fourier series

The Fourier series is a mathematical tool used to break down complex periodic functions into simpler sinusoidal components. It's essential in our discussion of string vibrations because it allows us to represent the complex, combined motion of the string as a sum of simpler sine and cosine terms. Here’s how it works in this context:

When a string vibrates, it doesn't do so in a simple, uniform motion. Instead, it forms standing waves that can be much more complicated. The displacement of the string can be described by a Fourier series, which is a sum of multiple sine and cosine functions:

\[ u(x, t) = \sum_{n=0}^\infty \frac{8 A}{\pi^2 (2n+1)^2} \sin \left( \frac{(2n+1) \pi x}{L} \right) \cos \left( \frac{(2n+1) \pi c t}{L} \right) \]
In this equation:

• \( n \) represents each mode of vibration or harmonic.
• \( \sin \left( \frac{(2n+1) \pi x}{L} \right) \) describes the spatial part for each mode.
• \( \cos \left( \frac{(2n+1) \pi c t}{L} \right) \) describes the time-dependent part.

The coefficients \( \frac{8 A}{\pi^2 (2n+1)^2} \) scale these functions based on the mode number \( n \).

By summing these sinusoidal components, the resulting series accurately represents the vibration of the string after being plucked. Each mode contributes to the overall shape and motion, creating the complex interplay of waves observed in the string’s dynamics.

Fourier series provide a powerful way of analyzing periodic functions and understanding wave patterns, which is crucial in fields like acoustics, signal processing, and even quantum mechanics.