How to solve this fields and waves question on coupled oscillators?

Question

Consider an infinite number of beads, each of mass 𝑚 and separated by distance ∆𝑥 , on a string along the 𝑥-axis and under tension 𝑇. Each normal mode eigenvector of the system has the 𝑖 mass experiencing displacement. The eigenvalue or 𝑡ℎ 𝑈 𝑖 = 𝐴𝑒 −𝑗(𝑖θ+ω𝑡) (angular) frequency ω of the normal mode is given by ω = 2 , 𝑇 𝑚∆𝑥 𝑠𝑖𝑛 θ ( 2 ) where the parameter θ can take on any value between 0 and π (other values merely duplicate existing modes).

a. What is the period λ of the spatial pattern of the normal mode in terms of θ? It may be helpful to write the position of the 𝑖 bead as 𝑥 = 𝑖(∆𝑥)

b. Now assume θ ≪ 1 and derive an expression for λ in terms of the frequency ω. What is the “wave velocity” of the normal mode pattern? This can be formed from the spatial period and frequency. You should also write it in terms of the mass density µ = 𝑚/∆𝑥.

c. What is the maximum normal mode frequency this system can have and what is the corresponding value of θ here? At this maximum frequency, what is the value of λ? Interpret this result.

Answer 1

Hey Jim, here is a solution of the parts below:

Part (a): Spatial Period lambda

Given:

U_i = A * e^(-j(i*theta + omega*t))

The position of the i-th bead is:

x_i = i * Delta_x

The spatial period lambda is the distance over which the displacement pattern repeats itself. For the pattern to repeat, we need:

U_(i+lambda/Delta_x) = U_i

This implies:

A * e^(-j((i + lambda/Delta_x)*theta + omega*t)) = A * e^(-j(i*theta + omega*t))

This simplifies to:

e^(-j((i + lambda/Delta_x)*theta)) = e^(-j(i*theta))

Since e^(-j*2*pi) = 1, we have:

- (lambda/Delta_x) * theta = 2*pi*n for any integer n

Choosing n = 1 for the fundamental period:

lambda = 2*pi*Delta_x / theta

Thus, the spatial period lambda in terms of theta is:

lambda = 2*pi*Delta_x / theta

Part (b): Small theta Approximation

Assume theta is much less than 1 (theta << 1):

omega = 2 * sqrt(T / (m * Delta_x)) * sin(theta/2)

For small theta:

sin(theta/2) is approximately equal to theta/2

Thus:

omega is approximately equal to 2 * sqrt(T / (m * Delta_x)) * (theta/2)

omega is approximately equal to theta * sqrt(T / (m * Delta_x))

Solving for theta:

theta is approximately equal to omega / sqrt(T / (m * Delta_x))

Using the earlier expression for lambda:

lambda is approximately equal to 2*pi*Delta_x / theta

lambda is approximately equal to 2*pi*Delta_x * sqrt(T / (m * Delta_x)) / omega

lambda is approximately equal to 2*pi * sqrt(T * Delta_x / m) / omega

The wave velocity v can be found using:

v = lambda * f

where f is the frequency f = omega / (2*pi)

Therefore:

v is approximately equal to 2*pi * sqrt(T * Delta_x / m) / omega * omega / (2*pi)

v is approximately equal to sqrt(T * Delta_x / m)

Expressing in terms of mass density mu = m / Delta_x:

v is approximately equal to sqrt(T / mu)

Part (c): Maximum Frequency

The maximum frequency occurs when theta = pi:

omega_max = 2 * sqrt(T / (m * Delta_x)) * sin(pi/2)

omega_max = 2 * sqrt(T / (m * Delta_x)) * 1

omega_max = 2 * sqrt(T / (m * Delta_x))

At this maximum frequency, theta = pi:

lambda = 2*pi * Delta_x / pi

lambda = 2 * Delta_x

**Interpretation**: The maximum normal mode frequency corresponds to the shortest wavelength, which is twice the separation between adjacent beads. This means that at the maximum frequency, each bead oscillates out of phase with its immediate neighbor, creating a highly localized and rapid oscillation pattern.

Summary

- (a) The spatial period lambda in terms of theta is lambda = 2*pi*Delta_x / theta.

- (b) For theta << 1, lambda is approximately equal to 2*pi * sqrt(T * Delta_x / m) / omega, and the wave velocity v is approximately equal to sqrt(T / mu).

- (c) The maximum frequency omega_max = 2 * sqrt(T / (m * Delta_x)), corresponding to theta = pi, and the spatial period at this frequency is lambda = 2 * Delta_x.

I would be happy to help you understand those further in tutoring sessions!

Best,

Andrew