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A spring of stiffness 0.3 N/mm is attached to a mass which has viscous damping device. When the mass was displaced and released, the period of vibration was 1.8 second and the ratio of consecutive amplitude was 4.2 : 1. The natural frequency of the system will be ______ (rad/s).
  • a)
    3.5
  • b)
    3.6
  • c)
    2.9
  • d)
    3.1
Correct answer is option 'B'. Can you explain this answer?
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To find the natural frequency of the system, we can use the formula:

Natural Frequency (ω) = (2π)/T

Where:
ω = Natural frequency
T = Period of vibration

Let's calculate the period of vibration first.

Given:
Ratio of consecutive amplitude = 4.2 : 1

The ratio of consecutive amplitudes can be written as:
A1/A2 = 4.2/1

Let's assume the initial amplitude is A1 and the amplitude after one complete cycle is A2.

A1/A2 = 4.2/1
A1 = 4.2A2

The period of vibration can be calculated using the formula:

T = (2π)/ω
T = Time for one complete cycle

Let's assume the time for one complete cycle is T1 and the time for the next complete cycle is T2.

T1/T2 = A1/A2 (Since the time period is inversely proportional to amplitude)
T1/T2 = 4.2A2/A2
T1/T2 = 4.2

We know that the ratio of consecutive periods is equal to the square root of the ratio of consecutive amplitudes.

T1/T2 = √(A1/A2)
4.2 = √(A1/A2)
4.2^2 = A1/A2
17.64 = A1/A2

Now, let's calculate the period of vibration.

T = T1 + T2

Substituting the value of T1/T2, we get:

T = 4.2T2 + T2
T = 5.2T2

Since the ratio of consecutive periods is 5.2, we can write:

T2/T = 1/5.2

Substituting the value of T, we get:

(1/5.2)T = (2π)/ω
T = (2πω)/5.2

Substituting the value of T from the given information, we get:

1.8 = (2πω)/5.2

Simplifying the equation, we find:

ω = (1.8*5.2)/(2π)
ω ≈ 3.6 rad/s

Therefore, the natural frequency of the system is approximately 3.6 rad/s.

Hence, the correct answer is option 'B'.
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A spring of stiffness 0.3 N/mm is attached to a mass which has viscous damping device. When the mass was displaced and released, the period of vibration was 1.8 second and the ratio of consecutive amplitude was 4.2 : 1. The natural frequency of the system will be ______ (rad/s).a)3.5b)3.6c)2.9d)3.1Correct answer is option 'B'. Can you explain this answer?
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