Lowest frequency that produces an interference maximum

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SUMMARY

The discussion centers on calculating the lowest frequency that produces an interference maximum at a microphone positioned between two speakers oscillating 180° out of phase, located 0.517 m apart. The microphone is 1.96 m from the midpoint of the speakers. The relevant equations include the path difference equation \( r_2 - r_1 = (n - 0.5)(\lambda) \) and the wave equation \( \lambda = \frac{v}{f} \), where \( v = 343 \, \text{m/s} \). The user initially struggled with determining the values for \( n \) to find the lowest and next lowest frequencies but later resolved their confusion.

PREREQUISITES
  • Understanding of wave interference principles
  • Familiarity with the equations of wave mechanics
  • Knowledge of the speed of sound in air (343 m/s)
  • Ability to manipulate algebraic equations for frequency and wavelength
NEXT STEPS
  • Study the concept of path difference in wave interference
  • Learn how to apply the wave equation \( \lambda = \frac{v}{f} \) in practical scenarios
  • Explore the effects of phase differences on interference patterns
  • Investigate higher-order interference maxima and their calculations
USEFUL FOR

Students in physics, particularly those studying wave mechanics and interference patterns, as well as educators looking for examples of practical applications of wave equations.

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Homework Statement



A microphone is located on the line connecting two speakers that are 0.517 m apart and oscillating 180° out of phase. The microphone is 1.96 m from the midpoint of the two speakers. What is the lowest frequency that produces an interference maximum at the microphone's location?

What is the next lowest frequency that produces an interference maximum at the microphone's location?




Homework Equations



r2 -r1 = (n-0.5)(lambda)
lambda=v/f
v=343m/s

The Attempt at a Solution



r2 and r1 are the distances from the speaker
so r2 is .517/2 + 1.96 = 2.2185
and r1 is .517/2 + 2.2185 = 2.477

i know i have to solve for frequency so i plugged (343m/s)/f into the equation

but then i got stuck. what do i plug in for n and how i change n for the lowest frequency and the next lowest?
 
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nevermind, figured it out

DISREGARD THE QUESTION!
 
Can you please tell me how to do this one? I've tried by using the equations but still can not figure it out.
Thanks!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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