Two speakers are connected to a signal source of 1000Hz and are set up 15 m apart. A person stands equidistant from each speaker so that each speaker individually produces the same intensity. When both speakers are on, the intensity is at a minimum. What could be the phase difference of the sound coming from the two speakers at this position?

0°

Explanation:

Since the speakers are equidistant from the person, there can only be constructive interference. . For constructive interference, the path difference ΔL = nλ a fractional number of wavelengths where λ = wavelength and n is a positive integer. For the intensity to be at a minimum, there must be destructive interference

λ = v/f where v = speed of sound in air = 343 m/s and f = frequency of signal = 1000 Hz

So, λ = v/f = 343 m/s ÷ 1000 Hz = 0.343 m

So for minimum intensity, n = 0

ΔL = nλ = (0)0.343 m = 0

So, the phase difference ΔФ = 2πΔL/λ  

= 2π(0 m)/0.343m

= 2π × 0

= 0°

there are no correct statementd on the list of choices you provided.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2