(1)A cowboy stands on horizontal ground between two parallel vertical cliffs. He

Question

(1)A cowboy stands on horizontal ground between two parallel vertical cliffs. He is not midway between the cliffs. He fires a shot and hears its echoes. The second echo arrives 2.04 s after the first and 1.69 s before the third. Consider only the sound traveling parallel to the ground and reflecting from the cliffs. (Take the speed of sound as 343 m/s.)

(a) What is the distance between the cliffs? m

(b) If he can hear a fourth echo, how long after the third echo does it arrive? s

(2)The intensity of a sound wave at a fixed distance from a speaker vibrating at 1.00 kHz is 0.700 W/m2.

(a) Determine the intensity if the frequency is increased to 2.30 kHz while a constant displacement amplitude is maintained.
W/m2

(b) Calculate the intensity if the frequency is reduced to 0.250 kHz and the displacement amplitude is quadrupled.
W/m2

Explanation / Answer

Given,

Speed of sound, v = 343 m/s

Let x1 be the cowboy’s distance from the nearer canyon wall and x2 be his distance from the farther cliff.

The sound for the first echo travels distance 2x1 .

For the second, 2x2 .

For the third, 2x1 + 2x2 ,

For the fourth echo, 2x1 + 2x2 + 2x1

then, 2x2 - 2x1 / 343 = 2.04 s

and 2x1 + 2x2 - 2x2 / 343 = 1.69

x1 = 0.5 x 343 x 1.69 = 289.84 m

2x2 / 343 = 2.04 + 1.69

x2 = 591.26

a) Distance between the cliffs, d = x1 + x2 = 881.1 m

b) time delay between third and fourth echo, t = 2x1+ 2x2 + 2x1 - ( 2x1 + 2x2 ) / 343 = 1.69 s

According to chegg guidelines I can answer one question at a time, so please post other question seperately. Please rate my answer if you find it helpful, good luck...

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