Answer to Question #122357 in Mechanics | Relativity for Lilly

Question #122357
A wave train moves can travel through 76 cm of a tight spring in 0.23 seconds, and it forms
3 crests in a distance of 24.0 cm. It then enters a loose spring and its speed decreases to
1.1 m/s. What is the wavelength in the loose spring? (Which property remains the same?)
Expert's answer

Explanations & Calculations

  • Velocity of the wave train within the first medium can be calculated to be,

"\\qquad\\qquad\n\\begin{aligned}\n\\small V_1 &= \\small \\frac{0.76 m }{0.23s} = \\bold{3.30ms^{-1}}\n\\end{aligned}"

  • Magnitude of the wavelength is defined as the distance between 2 identical points — one crest to another closest one, one trough to another closest one.
  • Therefore, 3 crests in a given distance means 2 wavelengths are generated within.

  • Therefore, the wavelength for the first medium ("\\lambda_1" ) could be calculated,

"\\qquad\\qquad\n\\begin{aligned}\n\u200d\\small 2\\lambda_1 &= \\small 0.24m\\\\\n\\small \\lambda_1 &= \\small 0.12 m\n\\end{aligned}"

  • For a wave propagates between several media, the frequency ("f" ) remains unchanged, only the wavelength changes due to the friction from the medium resulting a change in its velocity.

  • Therefore, for the wave within first medium,

"\\qquad\\qquad\n\\begin{aligned}\n\\small f &= \\small \\frac{3.3ms^{-1}}{0.12m} \\cdots(\\because V =f\\lambda)\n\\end{aligned}"

  • For the second situation,

"\\qquad\\qquad\n\\begin{aligned}\n\\small f &= \\small \\frac{1.1ms^{-1}}{\\lambda_2}\n\\end{aligned}"

  • Since the "f" is same,

"\\qquad\\qquad\n\\begin{aligned}\n\\small \\frac{3.3ms^{-1}}{0.12m} &= \\small \\frac{1.1ms^{-1}}{\\lambda_2}\\\\\n\\small \\lambda_2 &= \\small \\bold{0.04m=4cm}\n\\end{aligned}"

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