Question

When a 20-lb weight is suspended from a spring, the spring is stretched a distance of 1 in. Determine the natural frequency and the period of vibration for a 30-lb weight attached to the same spring

Answer 1

natural frequency = 2.55 Hz

period of vibration = 0.3915 s

Step-by-step explanation:

given data

weight = 20 lb

distance = 1 in =
`(1)/(12)` 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 =
`(weight)/(diatnace)` ..........1

k =
`(20)/(1/12)`

k = 240 lb/ft

so

frequency =
`\sqrt{(k)/(m) }` ..................2

put here value k and mass m =
`(weight)/(g)`

frequency =
`\sqrt{(240)/(30/32.2) }`

frequency = 16.05 rad/s

and

period of vibration =
`(2* \pi )/(frequency)`

period of vibration =
`(2* \pi )/(16.05)`

period of vibration = 0.3915 s

and

natural frequency =
`(1 )/(period of vibration)`

natural frequency =
`(1 )/(0.3915)`

natural frequency = 2.55 Hz