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In a spring-mass system, the mass is 0.1 kg and the stiffness of the spring is 1 kN/m. By introducing a damper, the frequency of oscillation is found to be 90% of the original value. What is the damping coefficient of the damper?
  • a)
    1.2 N.s/m
  • b)
    3.4 N.s/m
  • c)
    8.7 N.s/m
  • d)
    12.0 N.s/m
Correct answer is option 'C'. Can you explain this answer?
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Solution:

Given data:

Mass, m = 0.1 kg

Stiffness, k = 1 kN/m

Let the damping coefficient be c.

Let the natural frequency of the system be ω1.

Let the frequency of oscillation with damper be ω2.

Let the damping ratio be ζ.

Formulae:

Natural frequency of the system, ω1 = √(k/m)

Frequency of oscillation with damper, ω2 = ω1√(1-ζ²)

Damping ratio, ζ = c/(2√(mk))

Calculation:

Natural frequency of the system, ω1 = √(k/m) = √(1000 N/m / 0.1 kg) = 100 rad/s

Given that the frequency of oscillation with damper is 90% of the original value.

ω2 = 0.9ω1 = 0.9 x 100 = 90 rad/s

Substituting the values of ω1 and ω2 in the formula for damping ratio,

ζ = c/(2√(mk))

ζ = c/(2√(0.1 x 1000))

ζ = c/20

Substituting the value of ζ in the formula for ω2,

ω2 = ω1√(1-ζ²)

90 = 100√(1-ζ²)

1-ζ² = (9/10)²

1-ζ² = 81/100

ζ² = 19/100

ζ = 0.436

Substituting the value of ζ in the formula for damping ratio,

ζ = c/20

0.436 = c/20

c = 8.72 N.s/m

Therefore, the damping coefficient of the damper is 8.72 N.s/m.
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