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dimulka [17.4K]
3 years ago
8

A delicate telescope mirror and support that can be modelled as a mass-spring system with mass of 12 kg and support stiffness of

10,000 N/m and a damping ratio of 0.08 is set into motion with an initial displacement of 8 mm and initial velocity of 1 mm/s. Determine: The damped natural frequency of the system
Engineering
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:

damping natural frequency = 28.76 rad/s

Explanation:

given data

mass = 12 kg

stiffness = 10000 n/m

damping ratio = 0.08

displacement = 8 mm

initial velocity = 1 mm

to find out

damped natural frequency of the system

solution

we first find the natural frequency that is express as

natural frequency ω = \sqrt{\frac{k}{m} }    ..............1

here k is stiffness and m is mass

so ω = \sqrt{\frac{10000}{12} }

ω = 28.86 rad/s

so

damping frequency will be

damping frequency = ω × \sqrt{1- r^2}      .....................2

here r is damping ration

damping frequency = 28.86 × \sqrt{1- 0.08^2}

damping natural frequency = 28.76 rad/s

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