Given:

Mass of bob is 2.75 lb.

Length of pendulum is 24 in..

Mass of collar is 3 lb.

The relation of displacement is xc=δmsin(ωft).

The maximum amplitude of displacement is 0.4 in..

Frequency is 0.5 Hz.

Concept used:

Draw the diagram for system as shown below:

As the bob has a small displacement, from the above figure,

sinθ=(x−xc)l

Here, x is the final displacement of bob after the angular displacement, xc is the displacement and l is the length of pendulum.

Write the expression for the force balance in initial condition.

T=mg

Here, T is the tension in string, m is the mass of bob and g is the acceleration due to gravity.

Write the expression for Newton’s Law of motion in x-direction.

∑Fx=max

Substitute −Tsinθ for ∑Fx and x¨ for ax in the above expression.

−Tsinθ=mx¨ ...... (1)

Here, θ is the small angular displacement and x¨ is then acceleration in x-direction.

Substitute mg for T and (x−xc)l for sinθ in equation (1).

−mg(x−xc)l=mx¨−gl(x−xc)=x¨−gxl+gxcl=x¨

Substitute δmsin(ωft) for xc in the above expression and rearrange.

x¨+glx=glδmsin(ωft) ...... (2)

Write the general differential equation of motion.

x¨+ωn2x=F ...... (3)

Here, ωn is the natural circular frequency.

Compare equation (3) and (2).

ωn2=gl ...... (4)

Write the expression for the amplitude of forces vibration.

xm=δm1−(ωf2ωn2) ...... (5)

Here, xm is the amplitude of vibration, δm is the maximum amplitude of displacement and ωf is the forced frequency of vibration.

Calculation:

Substitute 32.2 ft/s2 for g and 24 in. for l in equation (4).

ωn2=32.224 in.(1 ft12 in.)=16.1 rad/s

Substitute 16.1 rad/s for ωn2 and 0.4 in. for δm and π for ωf in equation (5).

xm=(0.4 in.)1−(π216.1)=1.034 in.

The amplitude of motion of bob is 1.034 in..

Conclusion:

Thus, the amplitude of motion of bob is 1.034 in..

Concept used:

Draw the FBD of collar as shown below:

In the above Figure F is the force applied to maintain the motion, x¨c is the acceleration of collar and mc is the mass of collar.

Write the expression for Newton’s Law of motion in x-direction.

∑Fx=mcax

Substitute F+Tsinθ for ∑Fx and x¨c for ax in the above expression.

F+Tsinθ=mcx¨c

Substitute mg for T and (x−xc)l for sinθ in the above expression.

F+mg(x−xc)l=mcx¨c

Substitute ωn2 for gl in above equation and simplify for F.

F=−mωn2(x−xc)+mcx¨c ...... (6)

Write the relation for xc.

xc=δmsin(ωft)

Differentiate the above expression with respect to t.

x˙c=δmωf(cosωft)

Differentiate the above expression with respect to t.

x¨c=−δmωf2sin(ωft)

Substitute xmsin(ωft) for x, δmsin(ωft) for xc and −δmωf2sin(ωft) for x¨c in equation (6).

F=−mωn2((xmsin(ωft))−(δmsin(ωft)))+mc(−δmωf2sin(ωft))=−mωn2xmsin(ωft)+mωn2δmsin(ωft)−mcδmωf2sin(ωft)

Simplify the above expression for F.

`

F=(−mωn2xm+mωn2δm−mcδmωf2)sin(ωft) ...... (7)

Calculation:

Substitute 0.0854 lb-s2/ft for m, 16.1 rad/s for ωn2, 0.08613 ft for xm, 0.03333 ft for δm, 0.09316 lb-s2/ft for mc and π rad/s for ωf in equation (7).

F={−(0.0854)(16.1)(0.08613)+(0.0854)(16.1)(0.03333)−(0.09316)(0.03333)π2}sin(πt)=(−0.1184+0.0458−0.030)sin(πt)=−0.103sin(πt) lb

The periodic force required to maintain the motion of collar is −0.103sin(πt) lb.

Conclusion:

Thus, the periodic force required to maintain the motion of collar is −0.103sin(πt) lb.