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snow_tiger [21]
3 years ago
10

Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone

arranged as shown. For what fundamental frequency does their sound at the speakers produce constructive interference? The speed of sound is 340m/sec.

Physics
1 answer:
anygoal [31]3 years ago
7 0

Answer:

The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz

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Answer:

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6 0
3 years ago
Equations to use: v= λ ∙ f v=d/t
Margarita [4]

b. 460.8 m/s

Explanation:

The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

f=\frac{v}{2L}

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

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c. 18,000 m

Explanation:

The relationship between speed of the wave, distance travelled and time taken is

v=\frac{d}{t}

where

v = 6,000 m/s is the speed of the wave

d = ? is the distance travelled

t = 3 s is the time taken

Re-arranging the formula and substituting the numbers into it, we find:

d=vt=(6,000 m/s)(3 s)=18,000 m

3 0
3 years ago
Please. Physics is so difficult.
Softa [21]

Answer:

0.010 m

Explanation:

So the equation for a pendulum period is: y=2\pi\sqrt{\frac{L}{g}} where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

1.99 s = 2(3.14)\sqrt{\frac{L}{9.8 m\backslash s^2}}\\

Evaluate the multiplication in front

1.99 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}

Divide both sides by 6.28

0.317 s= \sqrt{\frac{L}{9.8 m\backslash s^2}}

Square both sides

0.100 s^2= \frac{L}{9.8 m\backslash s^2}

Multiply both sides by m/s^2  (the s^2 will cancel out)

0.984 m = L

Now now let's find the length when it's two seconds

2.00 s = 6.28\sqrt{\frac{L}{9.8m\backslash s^2}}

Divide both sides by 6.28

0.318 s = \sqrt{\frac{L}{9.8 m\backslash s^2}

Square both sides

0.101 s^2 = \frac{L}{9.8 m\backslash s^2}

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

0.994 m = L

So to find the difference you simply subtract

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4 0
1 year ago
Read 2 more answers
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dmitriy555 [2]

Answer:

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Explanation:

This is a pretty straightforward one.

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The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

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If the IBM does

138.6*10^12 = 1 second,

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x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

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