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Artyom0805 [142]
3 years ago
13

For a home sound system,two small speakers are located 52 cm closer to the listener than the other.for what value of the fundame

ntal frequency of audible sound will these speakers produce constructive interference at the listener? Speed of sound is 344 m/s
Physics
2 answers:
Iteru [2.4K]3 years ago
5 0

Answer:

did yyou finish the quiz?

Explanation:

BartSMP [9]3 years ago
3 0

Answer:

661.54 Hz

Explanation:

The lowest frequency (n=1) for the constructive interference is known as the fundamental frequency. For constructive interference, the difference in the distances of the speakers from the observer is equivalent to the multiple of the wavelength. Thus:

For fundamental frequency, the wavelength = 0.52/1  = 0.52 m

We know that: V = f*λ

V = speed of sound = 344 m/s

f = fundamental frequency = ?

λ = wavelength = 0.52 m

f = V/λ = 344/0.52 = 661.54 Hz

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4 years ago
Suppose the velocity of an electron in an atom is known to an accuracy of 2.0×103m/s (reasonably accurate compared with orbital
stiv31 [10]

Answer:

289.714 times bigger

Explanation:

\Delta x = Uncertainty in position

\Delta p = Uncertainty in momentum = \Delta v m

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From the Heisenberg uncertainty principle we have

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