Question

2.2 A mass of 0.50 kg hangs from the end of a light spring. The system is damped by a light sail attached to the mass so that the ratio of amplitudes of consecutive oscillations is equal to 0.90. It is found that 10 complete oscillations takes 25 s. Obtain a quantitative expression for the damping force and determine the damping factor y of the system.​

Answer 1

Answer:

Instructions: Write your answer in the question paper (question 1) Fill in the blanks: 1) The first natural number is 1 5q.units ti) Area of a square 2 is the only even prime number. iv) 47 86 tens + tenths * hundredth v) Perimeter of a rectangle: 2*___* ) units

Explanation:

Na ruko udati hun

Answer 2

Answer:

The expression of damping force is Fd = -0.00843v and the damping factor is y = 0.0169 sec⁻¹.

Explanation:

Given,

Mass, m = 0.5 kg

Time period, T = 25/2 = 12.5 sec

Amplitude, Af = 0.9A₀

The damping force is given by,

$F_{d} = -b[\frac{dx}{dt}]\\\\F_{d} = -b[v]\\\\and,\ \ b = \frac{2m}{T}ln[\frac{A_0}{A_f} ]$

b = (2×0.5/12.5)×ln(A₀/0.90A₀)

b = 0.08×ln(1.1111)

b = 0.00843 kg/sec

Therefore,

Fd = -0.00843v

The damping factor of the system will be

y = b/m

y = 0.00843/0.50

y = 0.0169 sec⁻¹

To learn more about damping force, click on the link below:

https://brainly.in/question/35245484

To learn more about the damping factor, click on the link below:

https://brainly.in/question/26284197

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