Let's break this question into parts and solve it step by step based on the given equation and information. The equation for the displacement of the rope is:

$y = (0.127 \, \text{m}) \sin \left( \frac{\pi x}{6.00} \right) \sin (13.0 \pi t),$

where:

• $x$ is in meters,
• $t$ is in seconds.

The rope is under a tension of $198 \, \text{N}$.

(a) The length of the rope

The standing wave equation indicates that this is the second harmonic. For the second harmonic, the rope has two complete loops, so the length of the rope $L$ is equal to the wavelength of the second harmonic:

$L = 2 \lambda,$

where $\lambda$ is the wavelength of one loop. From the spatial component of the wave equation, the argument of $\sin$ is:

$\frac{\pi x}{6.00}.$

This means that $\frac{\pi}{6.00}$ corresponds to $\frac{2\pi}{\lambda}$. Solving for $\lambda$:

$\frac{2\pi}{\lambda} = \frac{\pi}{6.00},$

$\lambda = 12.00 \, \text{m}.$

Thus, the length of the rope is:

$L = 2 \lambda = 2 \times 12.00 = 24.00 \, \text{m}.$

(b) The speed of the waves on the rope

The angular frequency $\omega$ is given as $13.0 \pi \, \text{rad/s}$, and the wave number $k$ is $\frac{\pi}{6.00}$. The wave speed $v$ is related to $\omega$ and $k$ by:

$v = \frac{\omega}{k}.$

Substitute $\omega = 13.0 \pi$ and $k = \frac{\pi}{6.00}$:

$v = \frac{13.0 \pi}{\pi / 6.00} = 13.0 \times 6.00 = 78.0 \, \text{m/s}.$

(c) The mass of the rope

The wave speed on a rope is also given by:

$v = \sqrt{\frac{T}{\mu}},$

where $T = 198 \, \text{N}$ is the tension and $\mu = \frac{m}{L}$ is the linear mass density of the rope. Solving for $m$:

$\mu = \frac{T}{v^2}, \quad m = \mu L.$

Substitute $T = 198 \, \text{N}$, $v = 78.0 \, \text{m/s}$, and $L = 24.00 \, \text{m}$:

$\mu = \frac{198}{(78.0)^2} = \frac{198}{6084} \approx 0.0325 \, \text{kg/m}.$

$m = \mu L = 0.0325 \times 24.00 \approx 0.780 \, \text{kg}.$

(d) The period of oscillation in the third harmonic

For the third harmonic, the rope will have three loops, and the wavelength becomes:

$\lambda_3 = \frac{L}{3}.$

The frequency $f$ of a standing wave is related to its speed and wavelength:

$f = \frac{v}{\lambda}.$

For the third harmonic:

$f_3 = \frac{v}{\lambda_3} = \frac{v}{L / 3} = \frac{3v}{L}.$

Substitute $v = 78.0 \, \text{m/s}$ and $L = 24.00 \, \text{m}$:

$f_3 = \frac{3 \times 78.0}{24.00} = \frac{234.0}{24.00} \approx 9.75 \, \text{Hz}.$

The period $T_3$ is the reciprocal of the frequency:

$T_3 = \frac{1}{f_3} = \frac{1}{9.75} \approx 0.103 \, \text{s}.$

Final Results:

(a) Length of the rope: $24.00 \, \text{m}$
(b) Speed of the waves: $78.0 \, \text{m/s}$
(c) Mass of the rope: $0.780 \, \text{kg}$
(d) Period of oscillation in the third harmonic: $0.103 \, \text{s}$

Would you like a detailed breakdown of any specific part? Here are 5 follow-up questions for practice:

1. How does the wavelength change for higher harmonics in a rope fixed at both ends?
2. Derive the formula for wave speed in terms of tension and mass density.
3. If the tension in the rope increases, how does the wave speed change?
4. What happens to the period of oscillation if the rope’s length is doubled?
5. How can you verify if a wave equation satisfies the wave equation mathematically?

Tip: Always ensure the boundary conditions of standing waves match the physical setup (e.g., nodes at fixed ends).