How does the frequency of a wave affect its power transmission?

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SUMMARY

The frequency of a wave significantly impacts its power transmission, as demonstrated in the discussion regarding two identical strings carrying sinusoidal waves. Wave A, with a frequency twice that of Wave B, transmits energy at a rate four times greater than Wave B. The power formula P = (1/2)μω²A²v indicates that power is proportional to the square of the angular frequency (ω), which is directly related to frequency. Thus, if the frequency doubles, the power increases by a factor of four, confirming that the correct answer is D.

PREREQUISITES
  • Understanding of wave mechanics and sinusoidal functions
  • Familiarity with the power formula for waves: P = (1/2)μω²A²v
  • Knowledge of angular frequency and its relationship to frequency (ω = 2πf)
  • Basic concepts of tension and mass per unit length in strings
NEXT STEPS
  • Study the relationship between frequency and power transmission in wave mechanics
  • Explore the derivation and application of the power formula for different wave types
  • Learn about the effects of tension and mass per unit length on wave speed and power
  • Investigate real-world applications of wave power transmission in engineering
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, as well as educators and professionals involved in acoustics and mechanical engineering.

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Homework Statement



Two identical but separate strings, with the same tension, carry sinusoidal waves with the same amplitude. Wave A has a frequency that is twice that of wave B and transmits energy at a rate that is ____ that of wave B.
A) half
B) twice
C) one fourth
D) four times
E) eight times

Homework Equations



[tex]P = \frac{1}{2}\mu\omega^{2}A^{2}v[/tex]
[tex]v = \lambda[/tex][tex]f[/tex]

The Attempt at a Solution



[tex]V_{B} = \lambda[/tex][tex]f_{B}[/tex]
[tex]V_{A} = \lambda[/tex][tex](2f_{B})[/tex] = [tex]2V_{B}[/tex]
[tex]P_{B} = \frac{1}{2}\mu\omega^{2}A^{2}v_{B}[/tex]
[tex]P_{A} = \frac{1}{2}\mu\omega^{2}A^{2}2v_{B}[/tex] = [tex]2P_{B}[/tex]

Thus my answer is B.

However, the answer key says D. What did I do wrong?
 
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If the strings are identical (same mu) and have the same tension, the velocity v is also identical. You can't assume lambda is the same, it can't be. What changes in your power formula is omega. How much does it change?
 
Ah okay, so w = 2pi / f.
Therefore the Pa = 4Pb.
Thank you, that makes sense.
 

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