Math Problem Statement

A rope, under a tension of 198 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.127 m) sin(πx / 6.00) sin(13.0πt), where x = 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, (c) the mass of the rope, and (d) if the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation?

Solution

Let's delve into the problem step by step. Here's how we tackle it:

(a) Length of the rope:

From the equation y=(0.127m)sin(πx/6.00)sin(13.0πt)y = (0.127 \, \text{m}) \sin(\pi x / 6.00) \sin(13.0 \pi t), we observe that sin(πx/6.00)\sin(\pi x / 6.00) corresponds to the standing wave function. The wavelength λ\lambda is twice the length of the rope for the second harmonic. Since λ=6.00m\lambda = 6.00 \, \text{m}, the rope's length is: L=λ2=6.002=3.00m.L = \frac{\lambda}{2} = \frac{6.00}{2} = 3.00 \, \text{m}.

(b) Speed of the waves:

The angular frequency ω=13.0π\omega = 13.0 \pi and the wave number k=π6.00k = \frac{\pi}{6.00}. Wave speed vv is given by: v=ωk=13.0ππ/6.00=13.06.00=78.0m/s.v = \frac{\omega}{k} = \frac{13.0 \pi}{\pi / 6.00} = 13.0 \cdot 6.00 = 78.0 \, \text{m/s}.

(c) Mass of the rope:

The tension T=198NT = 198 \, \text{N}, and wave speed vv relates to the mass per unit length μ\mu as: v=Tμ    μ=Tv2.v = \sqrt{\frac{T}{\mu}} \implies \mu = \frac{T}{v^2}. μ=198(78.0)2=1986084=0.03254kg/m.\mu = \frac{198}{(78.0)^2} = \frac{198}{6084} = 0.03254 \, \text{kg/m}. The total mass mm of the rope is: m=μL=0.032543.00=0.0976kg.m = \mu \cdot L = 0.03254 \cdot 3.00 = 0.0976 \, \text{kg}.

(d) Period of oscillation for the third harmonic:

For the third harmonic, the frequency f3=3f1f_3 = 3 \cdot f_1, where f1=v2Lf_1 = \frac{v}{2L}: f1=78.023.00=13.0Hz.f_1 = \frac{78.0}{2 \cdot 3.00} = 13.0 \, \text{Hz}. f3=313.0=39.0Hz.f_3 = 3 \cdot 13.0 = 39.0 \, \text{Hz}. The period TT is the reciprocal of frequency: T=1f3=139.0=0.02564s.T = \frac{1}{f_3} = \frac{1}{39.0} = 0.02564 \, \text{s}.

Final Answers:

(a) Length of rope: 3.00m3.00 \, \text{m}
(b) Speed of waves: 78.0m/s78.0 \, \text{m/s}
(c) Mass of rope: 0.0976kg0.0976 \, \text{kg}
(d) Period of oscillation: 0.02564s0.02564 \, \text{s}

Do you want further clarifications on any of these calculations?
Here are some related questions for deeper understanding:

  1. How is the wave equation derived for a standing wave?
  2. Why does the second harmonic imply half the wavelength fits the rope's length?
  3. What are the differences between harmonics in terms of wavelengths and frequencies?
  4. How does tension affect wave speed?
  5. What practical applications use these harmonic wave principles?

Tip: Always verify units in equations to ensure physical consistency!

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Math Problem Analysis

Mathematical Concepts

Wave motion
Standing waves
Second harmonic
Wave speed
Mass per unit length

Formulas

Wave speed: v = ω / k
Mass per unit length: μ = T / v²
Fundamental frequency: f = v / (2L)
Period: T = 1 / f

Theorems

Standing wave patterns on strings
Relationship between wave speed, tension, and mass density

Suitable Grade Level

Undergraduate Physics