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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

Respuesta :

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 = [tex]\sqrt{\frac{k}{m} }[/tex]    ..............1

here k is stiffness and m is mass

so ω = [tex]\sqrt{\frac{10000}{12} }[/tex]

ω = 28.86 rad/s

so

damping frequency will be

damping frequency = ω × [tex]\sqrt{1- r^2}[/tex]      .....................2

here r is damping ration

damping frequency = 28.86 × [tex]\sqrt{1- 0.08^2}[/tex]

damping natural frequency = 28.76 rad/s