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Luba_88 [7]
2 years ago
7

Two loudspeakers 6.0 m apart are playing the same frequency. If you stand 12.0 m in front of the plane of the speakers, centered

between them, you hear a sound of maximum intensity. As you walk parallel to the plane of the speakers, staying 12.0 m in front of them, you first hear a minimum of sound intensity when you are directly in front of one of the speakers.Part A)What is the frequency of the sound? Assume a sound speed of 340 m/s.I got 120.02 Hz. This is right, I checked.Part B)If you stay 12.0 m directly in front of one of the speakers, for what other frequencies between 100 Hz and 700 Hz is there a minimum sound intensity at this point?Express your answer numerically. If there is more than one answer, enter your answers in ascending order separated by commas.
Physics
1 answer:
Gnesinka [82]2 years ago
4 0

Answer:

The three frequencies between 100 and 700 Hz for minimum sound are

120.02 Hz, 360.06 Hz and 600.1 Hz

Explanation:

A) First of all, at the starting point centred between the speakers, as the sound is at a maximum, it means the speakers are in phase.

Thus,

When standing exactly in front of one speaker, you are 12 m from that speaker, and

√(6² + 12²) = 13.4164m from the other speaker.

Now,for minimum sound, the speakers must be out 180 degrees out of phase.

Thus; 13.4164 - 12 = 1.4164 m is an even multiple of half wavelengths of the sound.

The wavelength of sound is gotten from λ*F = V where λ is the wavelenth in m, F is the frequency in Hz, and V is the speed of sound in m/s.

So a half wavelength (H) = λ/2

So,

2H*F/(2n + 1) = V

F = (2n + 1)*V/(2H)

where n is an integer.

So,

The lowest frequency is when n = 0

Thus,

F = V/(2H)

F = 340/(2*1.4164) = 120.02 Hz

B) The next will be when n = 1

(2n + 1) = 3

F = 3 * 120.02 = 360.06 Hz

At n = 2

F = 5*120.02 = 600.10 Hz

Thus, The three frequencies between 100 and 700 Hz for minimum sound are

120.02 Hz, 360.06 Hz and 600.1 Hz

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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
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Answer:

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So electric field between plates will be equal to 1600\times 10^3KN/C

7 0
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