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A single degree of freedom spring mass system with viscous damping has a spring constant of 10 kN/m. The system is excited by a sinusoidal force of amplitude 100 N. If the damping factor (ratio) is 0.25, the amplitude of steady state oscillation at resonance is ________mm
    Correct answer is '20'. Can you explain this answer?
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    Hypothesis:
    We are given a single degree of freedom spring mass system with viscous damping. The system has a spring constant of 10kN/m and is excited by a sinusoidal force of amplitude 100 N. The damping factor (ratio) is 0.25. We need to find the amplitude of steady-state oscillation at resonance.

    Formula:
    The equation of motion for a single degree of freedom spring mass system with viscous damping can be written as:

    m*x'' + c*x' + k*x = F(t)

    Where,
    m = mass of the system (kg)
    x = displacement of the mass from equilibrium position (m)
    c = damping coefficient (N.s/m)
    k = spring constant (N/m)
    F(t) = external force (N)

    Analysis:

    Step 1: Calculation of Damping Coefficient (c)
    Given damping factor (ratio) = 0.25
    To calculate damping coefficient (c), we can use the formula:

    c = 2 * damping factor * sqrt(m * k)

    Given m = 1 kg and k = 10 kN/m, we can substitute these values to find c.

    c = 2 * 0.25 * sqrt(1 * 10 * 1000) = 2 * 0.25 * sqrt(10000) = 2 * 0.25 * 100 = 50 N.s/m

    Step 2: Calculation of Resonant Frequency (ωn)
    The resonant frequency (ωn) can be calculated using the formula:

    ωn = sqrt(k / m)

    Given k = 10 kN/m and m = 1 kg, we can substitute these values to find ωn.

    ωn = sqrt(10000 / 1) = sqrt(10000) = 100 rad/s

    Step 3: Calculation of Amplitude of Steady State Oscillation (Xss)
    At resonance, the amplitude of steady-state oscillation (Xss) can be calculated using the formula:

    Xss = (F(t) / k) / sqrt((1 - ω / ωn)^2 + (2 * damping factor * ω / ωn)^2)

    Given F(t) = 100 N and ω = ωn (as we are considering resonance), we can substitute these values to find Xss.

    Xss = (100 / 10000) / sqrt((1 - 1)^2 + (2 * 0.25 * 1)^2) = 0.01 / sqrt(0 + 0.25) = 0.01 / sqrt(0.25) = 0.01 / 0.5 = 0.02 m = 20 mm

    Therefore, the amplitude of steady-state oscillation at resonance is 20 mm.
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    A single degree of freedom spring mass system with viscous damping has a spring constant of 10kN/m. The system is excited by a sinusoidal force of amplitude 100 N. If the damping factor(ratio) is 0.25, the amplitude of steady state oscillation at resonance is ________mmCorrect answer is '20'. Can you explain this answer?
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    A single degree of freedom spring mass system with viscous damping has a spring constant of 10kN/m. The system is excited by a sinusoidal force of amplitude 100 N. If the damping factor(ratio) is 0.25, the amplitude of steady state oscillation at resonance is ________mmCorrect answer is '20'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A single degree of freedom spring mass system with viscous damping has a spring constant of 10kN/m. The system is excited by a sinusoidal force of amplitude 100 N. If the damping factor(ratio) is 0.25, the amplitude of steady state oscillation at resonance is ________mmCorrect answer is '20'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A single degree of freedom spring mass system with viscous damping has a spring constant of 10kN/m. The system is excited by a sinusoidal force of amplitude 100 N. If the damping factor(ratio) is 0.25, the amplitude of steady state oscillation at resonance is ________mmCorrect answer is '20'. Can you explain this answer?.
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