Question

Амплитуда затухающих колебаний убывает за десять колебаний на 1/10 часть своей первоначальной величины. Период колебаний 0,4 с. Определите логарифмический декремент затухания и коэффициент затухания.

Answer 1

N=10    A2=0.9*A1    T=0.4 c    δ=?  Θ=?
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t=N*T
δ*t=ln(A1/A2)
δ=ln(A1/A2)/(N*T)=ln(1/0.9)/(10*0.4)=0.026 c^-1
Θ=δ*T=0.026*0.4=0.01
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Answer 2

Calculating Logarithmic Decrement and Damping Coefficient

To calculate the logarithmic decrement of damping and the damping coefficient, we can use the given information about the amplitude of the decaying oscillations and the period of the oscillations.

Given: - Amplitude decreases by 1/10th over 10 oscillations. - Period of oscillations = 0.4 s.

Logarithmic Decrement of Damping

The logarithmic decrement of damping (δ) can be calculated using the formula: δ = ln(A/Aₙ) Where: - A = initial amplitude - Aₙ = amplitude after n periods

In this case, the amplitude decreases by 1/10th over 10 oscillations, so Aₙ = A/10.

Using the given period of oscillations (T = 0.4 s), we can calculate the number of periods (n) in 10 oscillations: n = 10 / T

Now, we can calculate the logarithmic decrement using the formula.

Damping Coefficient

The damping coefficient (ζ) can be calculated using the formula: ζ = δ / (2πn) Where: - δ = logarithmic decrement - n = number of periods

We can use the calculated value of δ and the number of periods to find the damping coefficient.

Let's calculate these values.