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katovenus [111]

3 years ago

10

Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but yo

u have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.

1 answer:

gtnhenbr [62]3 years ago

8 0

Answer:

a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg

Explanation:

In a simple harmonic movement with friction, we can assume that this is provided by the speed

fr = -c v

when solving the system the angular value remains

w² = w₀² + (c / 2m)²

They give two conditions

1) m₁ = 1 kg

f₁ = 1.1 Hz

the angular velocity is related to frequency

w = 2π f₁

Let's find the angular velocity without friction is

w₂ = k / m₁

we substitute

(2π f₁)² = k / m₁ + (c / 2m₁)²

2) m₂ = 2 kg

f₂ = 0.8 Hz

(2π f₂)² = k / m₂ + (c / 2m₂)²

we have a system of two equations with two unknowns, so we can solve it

we solve (c / 2m)² is we equalize the expression

(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁

k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)

k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)

a) Let's calculate

k = 4 π² (0.8² -1.1²) / (½ -1/1)

k = 39.4784 (1.21) / (-0.5)

k = 95.54 N / m

now we can find the constant of friction

(2π f₁) 2 = k / m₁ + (c / 2m₁)²

c2 = ((2π f₁)² - k / m₁) 4m₁²

c2 = (4ππ² f₁² - k / m₁) 4 m₁²

let's calculate

c² = (4π² 1,1² - 95,54 / 1) 4 1²

c² = (47.768885 - 95.54) 8

c² = -382.1689

c = 19.55

b) f₃ = 0.2 Hz

m₃ =?

(2πf₃)² = k / m₃ + (c / 2m₃) 2

we substitute the values

(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²

1.579 = 95.54 / m₃ + 95.542225 / m₃²

let's call

x = 1 / m₃

x² = 1 / m₃²

- 1.579 + 95.54 x + 95.542225 x² = 0

60.5080 x² + 60.5080 x -1 = 0

x² + x - 1.65 10⁻² = 0

x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2

x = [1 ± 1.03] / 2

x₁ = 1.015 kg

x₂ = -0.015 kg

Since the mass must be positive we eliminate the second results

x₁ = 1 / m₃

m₃ = 1 / x₁

m₃ = 1 / 1.1015

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Answer:

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b) See the proof below

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Explanation:

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Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

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$k = -\frac{ln(2)}{T}$

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And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

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For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

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Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

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