Intensity of two sound pressure waves

AI Thread Summary
The intensity of sound pressure waves is directly proportional to the square of the amplitude and frequency, while inversely proportional to the square of the radius. The discussion clarifies that the areas mentioned are not relevant when comparing the intensities of two different sound sources at the same point in space and time. It emphasizes that the comparison should focus on the readings from detectors positioned at equal distances from the sources. Ultimately, it concludes that one wave has greater intensity than the other, confirming the understanding of the relationship between amplitude and intensity. This highlights the importance of focusing on detector readings rather than the areas of the waves.
iQadmat
Messages
3
Reaction score
3
Homework Statement
Ans: C

I tried to use the intensity formulae and after manipulating it got an expression which gave me wrong answer.
It seems the areas are not equivalent somehow, any explanation?
Relevant Equations
I = P/A
1658379625863.png

1658379637746.png
 
Physics news on Phys.org
Hi @iQadmat and welcome to PF.

How does the intensity depend on the amplitude? You can read the amplitudes of the two waves off the graph and compare.
 
kuruman said:
Hi @iQadmat and welcome to PF.

How does the intensity depend on the amplitude? You can read the amplitudes of the two waves off the graph and compare.
Hi and thanks for welcoming me!

I know the intensity is directly proportional to amplitude^2 frequency^2 and inversely proportional to radius^2, but my point was the areas (which I showed in ss) are different somehow and I want to understand the differences between them and what they represent.

So far my understanding*
[The initial 'A' is showing the area at which the power from source is acting upon but the final 'A' (later derived from 'F') is not the initial 'A' but is the area of the wave particle which times by pressure of wave equals 'F' .]
 
The areas are irrelevant. This is a comparison of intensities of two different sources at the same point in space and time. If you must, assume that the two sources are at the same distance from two detectors with identical areas. Compare the detectors' readings. Clearly, wave 2 has greater intensity than wave 1. By what factor?
 
kuruman said:
The areas are irrelevant. This is a comparison of intensities of two different sources at the same point in space and time. If you must, assume that the two sources are at the same distance from two detectors with identical areas. Compare the detectors' readings. Clearly, wave 2 has greater intensity than wave 1. By what factor?
I got it now, Thanks for your help!
 
  • Like
Likes Delta2 and kuruman
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