Two loudspeakers at an outdoor rock concert are located 3.8 meters apart. you are standing 17.1 meters from one of the speakers and 20.3 from the other. during a sound check, the technician sends the exact same frequency to both speakers while you listen. the technician starts at 20hz and slowly increases it to 30,000hz.

a) what is the lowest frequency where you will hear a minimum signal ?
f = hz
b) what is the second lowest frequency where you will hear a minimum signal ? f = hz
c) what is the lowest frequency where you will hear a maximum signal ? f = hz
d) what is the second lowest frequency where you will hear a maximum signal ? f = hz

Explanation:

Path difference of sound wave = 20.3 - 17.1 m

= 3.2 m

For minimum signal , there should be destructive interference of sound wave .

For it , Path difference should be odd multiple of wavelength of sound / 2

or

3.2 = (2n+1)λ / 2

λ = 6.4 /(2n+1)

At 20 Hertz , wavelength

= 340/20x 2

= 17 m

At 30000 Hertz , wavelength

= .0113 m  

The range of wavelength we can take up is .0113 m  to  17 m

λ = 6.4 /(2n+1)

a ) For lowest frequency , wavelength should be highest. λ is highest when

n is lowest . If n = 1

λ = 6.4 /3

= 2.133 which lies in the range so lowest frequency

= 340 / 2.133

= 160 Hz.

b )

Second lowest frequency

n = 2

λ = 6.4 /(2n+1)

= 6.4 / 5

= 1.28 m

frequency

= 340 / 1.28

= 266 Hz

c ) For maximum sound

Path difference should be  multiple of wavelength of sound

or 3.2 = nλ

 λ = 3.2 /n

For lowest frequency ,   λ should be maximum and n should be minimum .

Minimum n = 1

λ = 3.2

Frequency

= 340 / 3.2

= 106 Hz

d ) For second lowest

n =2

λ = 3.2 /n

λ = 3.2 /2

= 1.6 m

frequency

= 340 / 1.6

= 212 Hz

b) phosphorus pentoxide