"Diffraction: Why Bass Music is Easier to Hear

  • Thread starter Thread starter student07
  • Start date Start date
  • Tags Tags
    Sound Waves
AI Thread Summary
Bass music is easier to hear from a distance due to its longer wavelengths, which allow it to diffract more effectively than higher treble frequencies. The lower frequency of bass notes enables them to travel through obstacles like doorways with less attenuation. While the primary factor is frequency, the discussion raises questions about other potential influences on sound propagation. Understanding diffraction helps clarify why bass sounds are more prominent in distant listening scenarios. Overall, the physics of sound waves explains the dominance of bass in such contexts.
student07
Messages
36
Reaction score
1

Homework Statement


If you hear music in the distance, would you be more likely to hear the treble or the bass notes of the music? Using your knowledge of diffraction, explain your answer.

2. The attempt at a solution
So I found that the bass's waves are carried at a longer distance than the treble ones and can pass through doorways or small openings at a greater distance than that of the treble since their frequency is not affected a lot by diffraction. Thus, the wavelength of the bass carries a longer distance than that of the treble but is it only because of the frequency?
 
Physics news on Phys.org
Since the only difference between bass and treble is frequency what else could it be?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top