The small block at A has a mass of 4 kg and is mounted on th

The small block at A has a mass of 4 kg and is mounted on the bent rod having negligible mass. If the rotor at B causes a harmonic movement delta_B = (0.1 cos 12t) m. where t is in seconds, determine the steady-state amplitude of vibration of the block. (Figure 1) Express your answer with the appropriate units.

Solution

>> Due to harmonic vibration of, y = 0.1*cos(12t), Spring will extend, resulting in spring force at Pt.C (tip of spring) .

So, Fs = Spring Force Acting = ky = 15*0.1*cos(12t) = 1.5*cos(12t)

>> Now, Considering the bent rod, AOC, forces acting are:

1). Wa = Weight of block at A = 4*9.81 = 39.24 N

2). Fs = Spring Force at C = 1.5*cos(12t)

>> As, Maximum Amplitude of Vibration at C = 0.1 m

and, if C moves by 0.1 m, lets assume A moves by \'x\'

From Geometry,

0.1/1.2 = x/0.6

=> x = 0.05 m .........REQUIRED AMPLITUDE OF VIBRATIONS OF BLOCK .......ANSWER.....