Answer to Question #148864 in Mechanics | Relativity for Gaurav

Question #148864
Derive the speed of transverse waves on a uniform string, with the help of suitable diagram.
1
Expert's answer
2020-12-06T17:21:15-0500

Explanations & Calculations


  • For an any transverse wave of the form "\\small" "\\small y(x,t)=f(x \\pm vt)", the derivative form can be written as "\\frac{\\partial^2y}{\\partial x^2}\\small(x,t)=\\frac{1}{v^2} \\frac{\\partial^2y}{\\partial^2 t^2}\\small(x,t) \\cdots (1)"





  • Consider an elemental segment (AB) of a uniform thread taut to a tension of T and total mass of M. (First figure)
  • Mass of the elemental segment is "\\small \\Delta M = \\frac{M}{L}\\times \\Delta x"
  • The quantity "\\small \\frac{M}{L}" is known as the linear density m
  • Then consider it after plucked & has entered into a wavy motion. (second figure) Here the elemental segment is located in a slanted position between AB.
  • Then the thread experiences a net tension ("\\small T_{net}" ) which comprises of the previous tension and the restoring force which acts to reverse the change made in to the system. And it has the usual horizontal & vertical components.
  • Since the segment do not travel horizontally, those vertical components remains equal & cancels each other.
  • And for the vertical components apply Newton's second law upwards

"\\qquad\\qquad\n\\begin{aligned}\n\\small F_1 -F_2 &= \\small \\Delta m a=\\Delta m \\frac{\\partial^2 y}{\\partial t^2}\\\\\n\\small F_1 -F_2 &= \\small m \\Delta x \\frac{\\partial^2 y}{\\partial t^2 }\\cdots (1)\n\\end{aligned}"

  • From the wave form above,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\tan\\theta _1 =\\small \\frac{F_1}{F_x} \n\\end{aligned}" and "\\qquad\\qquad\n\\begin{aligned}\n\\small \\tan\\theta_2 =\\frac{F_2 }{F_x}\n\\end{aligned}"

  • These tan values are the gradients of the waveform at those given points. Therefore,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\tan \\theta _1 &= \\small \\frac{\\partial y}{\\partial x}\\Big|_B\n\\end{aligned}" and "\\qquad\\qquad\n\\begin{aligned}\n\\small \\tan\\theta_2 =\\small\\frac{\\partial y}{\\partial x}\\Big|_A\n\\end{aligned}"


  • Substituting these in (1),

"\\qquad\\qquad\n\\begin{aligned}\n\\small F_x\\Bigg[\\frac{\\partial y}{\\partial x}\\Big|_B-\\frac{\\partial y}{\\partial x}\\Big|_A\\Bigg]&= \\small m\\Delta x\\frac{\\partial^2y}{\\partial t^2}\\\\\n\\small \\frac{\\Bigg[\\frac{\\partial y}{\\partial x}\\Big|_B-\\frac{\\partial y}{\\partial x}\\Big|_A\\Bigg]}{\\Delta x}&= \\small \\frac{m}{F_x}\\frac{\\partial^2y}{\\partial t^2}\n\\end{aligned}"

  • And taking limit of this equation as "\\small \\lim_{\\Delta x\\to0}" ,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\frac{\\partial^2y}{\\partial x^2 } &= \\small \\frac{m}{F_x}\\frac{\\partial ^2 y}{\\partial t^2 }\\cdots(2)\n\\end{aligned}"

  • Comparing (2) with (1),

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\frac{m}{F_x} &= \\small \\frac{1}{v^2}\\\\\n\\small |v| = \\sqrt{\\frac{F_x}{m}}\n\\end{aligned}"

  • This horizontal component is the tension of the string ,therefore,

"\\qquad\\qquad\n\\begin{aligned}\n\\small |v| = \\sqrt{\\frac{T}{m}}\n\\end{aligned}"


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS