Question

Physics - Do both problems and both parts to each and I'll be sure to give you an upvote. Thanks

1) A concert loudspeaker suspended high above the ground emits 45 W of sound power. A small microphone with a 1.0 cm2 area is 75 m from the speaker a) What is the sound intensity at the position of the microphone? b) How much sound energy impinges on the microphone each second? 2) An opera singer in a convertible sings a note at 600 Hz while cruising down the highway at 90 km/h. What is the frequency heard by a) b) A person standing beside the road in front of the car? A person on the ground behind the car?

Answer 1

1)

(a) I = P/4pi*r²

= 45/(4pi*75²)

= 6.67 x 10^-4 W/m²

(b)

1 watt = 1joule /1 second

Energy incident per m^2 per second is intensity I

Energy incident per 1.0x10^-4m^2 per second is

I x 1.0x10^-4 = 6.67x10^-4x1.0x10^-4 = 6.366 x 10^-8 J

2)

(a)

90km/hr = 90/3600 x 1000 m/s = 25m/s

speed of sound = 330 m/s

frequency is heard higher infront of the car. velocity of sound heard is in effect 330 + 25 = 355m/s

f = 600Hz x (355) / 330 = 645.4545454545fourfive hertz

(b)

for the person behind the car, the frequency is lower. the speed of the sound wave relative to that person is speed of sound minus speed of car

so f = 600 x (330-25)/330 = 554.545454.....Hz

Answer 2

1)

(a) I = P/4pi*r²

= 45/(4pi*75²)

= 6.67 x 10^-4 W/m²

(b)

1 watt = 1joule /1 second

Energy incident per m^2 per second is intensity I

Energy incident per 1.0x10^-4m^2 per second is

I x 1.0x10^-4 = 6.67x10^-4x1.0x10^-4 = 6.366 x 10^-8 J

2)

(a)

90km/hr = 90/3600 x 1000 m/s = 25m/s

speed of sound = 330 m/s

frequency is heard higher infront of the car. velocity of sound heard is in effect 330 + 25 = 355m/s

f = 600Hz x (355) / 330 = 645.4545454545fourfive hertz

(b)

for the person behind the car, the frequency is lower. the speed of the sound wave relative to that person is speed of sound minus speed of car

so f = 600 x (330-25)/330 = 554.545454.....Hz