When you double your distance from a sound source

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The discussion centers on a test question about calculating the distance to a cliff based on the time taken for an echo to return. The correct calculation shows that the distance to the cliff is 662 meters, as the total distance includes the sound traveling to the cliff and back. Additionally, the intensity of sound decreases inversely with the square of the distance, meaning that doubling the distance results in the intensity being reduced to one-fourth of its original value. The initial confusion arose from misunderstanding the total distance in the echo calculation and the relationship between distance and sound intensity. The correct answers are 662 meters for the cliff distance and 1/4 for the sound intensity reduction.
Dx
This can't be right.

I missed this test question here.

You shout at a cliff and hear the echo in 4s. the temp is 0 degrees C. How far away is the cliff?

I think the answers were a)662 b) 680 c) 1320 d)1760

I chose c because

t=d/s or d=t*s

now, i substitued d = 4s * 331m/s = 1324

why did i miss this? I am pretty sure the formula and math is
right?


also one more question



when you double your distance from a sound source radiating in all directoins. what happens to the intensity of the sound? is reduces to 1/2 its orginal value.

Thats what i said unless its 1/4. what is the right answer.

Dx :wink:
 
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a)
You computed:
d = 4s * 331m/s = 1324

That was exactly correct -- for the distance from you to the cliff and back. D is twice the distance from you to the cliff. So, the correct answer is a)662 m.

b)
Sound intensity varies inversely with the square of the distance, so the correct answer to the second question was 1/4.
 
thanks gnome!
 
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