When a certain motor is started, it is noticed that its supporting frame begins to resonate when the motor speed passes through 900 rpm. At the operating speed of 1750 rpm, the support oscillates with an amplitude of 8 mm. Determine the amplitude that would result at 1750 rpm if the support were replaced with one having one-half the stiffness.

X₂     = 6.74 mm

Explanation:

As we know that:

ω = 900 (2π/60) rad/ sec = natural frequency

ω    = 1750(2π/60) rad/ sec = maximumm frequency

Also,

r ₁ = ω / ω = [1750(2π/60)]/[900(2π/60)] = 1.94

Now we know that general formula for stiffness is

ω = ---------- equation (1)

Similarly

ω = ---------- equation (2)

put k = k/2 in equation 2, we get

ω =

Now, by equating equation 1 and 2, we get

ω = ω

  or

(1/)ω = ω

By putting ω = 900 (2π/60) we get

(1/)900 (2π/60) = ω

By solving it we get

ω = 636.4(2π/60)

Also we know that

r ₁ = ω / ω

By putting values in it, we get

r₂ = [1750(2π/60)] / [636.4 (2π/60)]

r₂ = 2.75

For τ = 0, we have rotating unbalanced equation as

X =[m∈/m][r²/(1-r²)]

For X=8, r = r₁ ;  8 =[m∈/m][r₁²/(1-r₁²)]    ---------------equation (a)

For X=X₂, r = r₂ ;  8 =[m∈/m][r₂²/(1-r₂²)] ---------------equation (b)

Dividing equation a by b , we get

X₂/8 = (r₂/r₁)²[(1-r₁²)/(1-r₂²)]

Now, put r₁ = 1.94 ; r₂ = 2.75

we get

X₂/8 = (2.75/1.94)²[(1-1.94²)/(1-2.75²)]

        = 0.842

X₂     = 6.74 mm

The amplitude of one half thickness is calculated to be 7.74mm