two speakers C & D are driven in step at 600 Hz by the same audio oscillator. these speakers both start out at 5.0 m from the listener, but speaker C is slowly moved away. if C keeps moving, at what distance d will the speakers produce the second destructive interference at the listener? the speed of sound is 340 m/s.

A) 0.97 m

B) 0.86 m

C) 0.45 m

D) 0.28 m

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wagonbelleville wagonbelleville · Answer 1 · 2019-12-13T09:28:35+01:00

Answer:

option B

Explanation:

given,

frequency of the speaker = 600 Hz

speed of sound = 340 m/s

wavelength = ?

[tex]\lambda = \dfrac{v}{f}[/tex]

[tex]\lambda = \dfrac{340}{600}[/tex]

[tex]\lambda = 0.567 m[/tex]

for second destructive interference,

the path difference,

[tex]d =(2k-1)\dfrac{\lambda}{2}[/tex]

k = 2

[tex]d =(2\times 2-1)\dfrac{\lambda}{2}[/tex]

[tex]d =\dfrac{3\lambda}{2}[/tex]

[tex]d =\dfrac{3\times 0.567}{2}[/tex]

d = 0.86 m

the correct answer is option B

ibiscollingwood ibiscollingwood · Answer 2 · 2022-02-01T14:56:48+01:00

The speakers produce the second destructive interference at the distance of 0.86 meter.

Option B is correct.

The wavelength is given as,

[tex]\lambda=\frac{v}{f} =\frac{340}{600} =0.567m[/tex]

For Destructive interference, distance is given as,

[tex]d=(2k-1)*\frac{\lambda}{2}[/tex]

For second destructive interference , substitute k = 2.

[tex]d=(4-1)*\frac{0.567}{2}=\frac{1.701}{2} =0.86m[/tex]

Thus, the speakers produce the second destructive interference at the distance of 0.86 meter.

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