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2 years ago

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The G string on a guitar is 69 cm long and has a fundamental frequency of 196 Hz. A guitarist can play different notes by pushin

g the string against various frets, which changes the string's length. The third fret from the neck gives B♭ (233.08 Hz); the fourth fret gives B (246.94 Hz).
How far apart are the third and fourth frets?

Express your answer to two significant figures and include the appropriate units.

1 answer:

Readme [11.4K]2 years ago

6 0

Answer:

Δd = 23 cm

Explanation:

When the eta string on the guitar has nodes at its ends, so the waves produced give rise to a standing wave, which can be described by the following expressions

Fundamental L = ½ λ

1st harmonic L = 2 ( λ / 2)

2nd harmonic L = 3 ( λ / 2)

Harmonic n L = n λ / 2

Where n is an integer

The speed of the rope is given by the ratio

v = λ f

This speed is constant since it depends on the tension and the linear density of the rope

Let's calculate the speed with the initial data

v = 0.69 196

v = 135.24 m / s

Let's look for the wavelength for the two frequencies

λ₁ = v / f₁

λ₁ = 135.24 / 233.08

λ₁ = 0.58022 m

λ₂ = v / f₂

λ₂ = 135.24 / 246.94

λ₂ = 0.54766 m

Let's replace in the resonance equation

Lₙ = n λ/2

For the third fret

m = 3

L₃ = 3 0.58022 / 2

L₃ = 0.87033 m

For the fourth fret

m = 4

L₄ = 4 0.54766 / 2

L₄ = 1.09532 m

The distance between the two frets is

Δd = L₄ –L₃

Δd = 1.09532 - 0.87033

Δd = 0.22499 m

Δd = 22.5 cm = 23 cm

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Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

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Then:

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Now;Initially the volume of the glycerin before the ice cube starts to melt is:

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The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

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$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

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Refer to the diagram shown below.

Define unit vectors along the x and y axes as respectively $\hat{i} \, and \, \hat{j}.$

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Answer:

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We are given;

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d2 - d'2 = 75(656.3) - 57(434)

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<h3>PROOF:</h3>

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