SOUND AND WAVES. AM and FM radio waves are transverse waves that consist of electric and magnetic disturbances. These waves travel at a speed of 3* 10.

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Presentation transcript:

SOUND AND WAVES

AM and FM radio waves are transverse waves that consist of electric and magnetic disturbances. These waves travel at a speed of 3* 10 ^ 8 m/s. A station broadcast on AM radio waves whose frequency is 1230*10^3 Hz and an FM radio wave whose frequency is 91.1 * 10^6. Find the distance between adjacent crests in each wave?

AM: Wavelength = v/f = 3 * 10^8 / 1230 * 10^3 = 244 m FM Wavelength = v/f = 3 * 10^8 / 91.9 * 10 ^6= 3.26 m

Transverse waves travel on the strings of an electric guitar after the strings are plucked. The length of each string between its two fixed ends is.628 m and the mass is.208 g for the highest pitched E string and 3.32 g for the lowest pitched E string. Each String is under a tension of 226 N. Find the speeds of the waves on the two strings.

V= (F/ m * L)^.5 = (266/.208*10^-3 *.628)^.5 = 826 m/s for the highest pitch V= (F/ m * L)^.5 = (266/ 3.32*10^-3 *.628)^.5 = 207 m/s for the lowst pitch

12*10^5 W of sound power passes perpendicularly through the surfcaes. These surfaces have areas of A1=4 m^2 and A2=12 m^2. Determine the sound intensity of each surface and discuss why listener 2 hears a quieter sound than listener 1.

I= P/Av= 12*10^-5 / 4 = 3*10^-5 I= P/Av= 12*10^-5 / 12 = 1*10^-5 The reason listener 2 hears a quieter sound than listener 1 is because the same amount of sound is spread over a large amount of space.

A rocket explodes high in the air. Assume that the sound spreads out uniformly in all directions and that reflections from the ground can be ignored. When the sound reaches listener 2, who is r2=6540 m away from the explosion, the sound has an intensity I2 of.10 W/m^2. Why is the sound intensity detected by listener 1, who is r1= 160 meters away from the explosion?

I= P / 4 pie r^2 I1 / I2 = (P / 4 pie r^2) / (P / 4 pie r^2 ) Simply to I1 / I2 = r^2 / R^2 16*.1 / 16 = 1.6 W/m^2

Audio system 1 produces an intensity level of B1=90 dB, and system 2 produces an intensity level of B2= 93 dB. The corresponding intensities are I1 and I2. Determine the ration I2/I1

93 – 90 = 10 log ( I2 / I1) 3/10 = log ( I2 / I1).3 =log ( I2 / I1)  10^.3 = ( I2 / I1) = 2

A high-speed train is traveling at a speed of 44.7,/s when the engineer sounds the 415 Hz warning horn. The speed of sound is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a person standing at a crossing, when the train is (a) approaching and (b) leaving?

A. Fo = fs ( 1/ 1- Vs/v) = 415 ( 1/ ( 1- ( 44.7 / 343)) = 477 Hz A. Wavelength = v/ fo = 343 / 477 =.719 meters B. Fo = fs ( 1/ 1- Vs/v) = 415 ( 1/ ( 1 + ( 44.7 / 343)) = 367 Hz Wavelength = v/ fo = 343 / 367 =.935 meters