Destructive interference Problem

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SUMMARY

The problem involves two point sources, A and B, that are 6.0 cm apart, generating sinusoidal waves in phase. The points of destructive interference are located at distances of 0.6 cm and 2.2 cm from source A. The wavelength of the water waves is determined to be 3.2 cm, with a total of four points of destructive interference along the line AB, specifically at distances of 0.6 cm, 2.2 cm, 3.8 cm, and 5.4 cm from source A.

PREREQUISITES
  • Understanding of wave interference principles
  • Knowledge of the superposition principle in wave mechanics
  • Familiarity with sinusoidal wave equations
  • Basic skills in calculating path differences in wave phenomena
NEXT STEPS
  • Study the concept of wave interference patterns in detail
  • Learn about the superposition principle and its applications in physics
  • Explore the derivation and implications of the sinusoidal wave equation
  • Investigate the effects of varying distances between wave sources on interference patterns
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Students studying wave mechanics, physics educators, and anyone interested in understanding wave interference phenomena in practical scenarios.

patrickmoloney
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Homework Statement



two point sources A and B generate sinusoidal waves in a water surface. The sources are 6.0 cm apart and vibrate in phase with the same amplitude and frequency. Points of destructive interference are observed on AB, the line joining the two sources. the two nearest such points to source A occur 0.6 cm and 2.2 cm from it. determine the wavelength of the water waves. How many points of destructive interference are there in total along the line AB

Homework Equations



y1= Asin(kx1-ωt+ [itex]\Phi[/itex] )

y2= Asin(kx2-ωt+ [itex]\Phi[/itex] )

Superposition Principle : y= y1+y2

y= 2A[sin(kx1+kx2/2-ωt+ [itex]\Phi[/itex] )cos(kx1-kx2/2)

The Attempt at a Solution



Have no idea. I've been at it for ages..my attempts have conjured up nothing.
 
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From a PM exchange -- Patrick has made progress on this question...

patrickmoloney said:
I know that path difference between 2 nodes is one wavelength, perhaps if A goes to b you increase distance by 1.6 cm? of and then decrease B from A by 1.6 cm. Then add these two distances and that is the wavelength? Thanks a million! You helped me solve it. Wavelength is 3.2 cm with 4 points of destructive interference, of course! How could I have been so dumb.. 0.6, 2.2, 3.8 and 5.4 ...it all makes sense now. Thank you again.
 
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