Respuesta :
Answer:
natural frequency = 2.55 Hz
period of vibration = 0.3915 s
Explanation:
given data
weight = 20 lb
distance = 1 in = [tex]\frac{1}{12}[/tex] ft
weight = 30 lb
to find out
Determine the natural frequency and the period of vibration
solution
we first calculate here stiffness k by given formula that is
k = [tex]\frac{weight}{diatnace}[/tex] ..........1
k = [tex]\frac{20}{1/12}[/tex]
k = 240 lb/ft
so
frequency = [tex]\sqrt{\frac{k}{m} }[/tex] ..................2
put here value k and mass m = [tex]\frac{weight}{g}[/tex]
frequency = [tex]\sqrt{\frac{240}{30/32.2} }[/tex]
frequency = 16.05 rad/s
and
period of vibration = [tex]\frac{2* \pi }{frequency}[/tex]
period of vibration = [tex]\frac{2* \pi }{16.05}[/tex]
period of vibration = 0.3915 s
and
natural frequency = [tex]\frac{1 }{period of vibration}[/tex]
natural frequency = [tex]\frac{1 }{0.3915}[/tex]
natural frequency = 2.55 Hz
