For the system shown the amplitude of vibration decays to ha

For the system shown, the amplitude of vibration decays to half of its initial value in 11 cycles with a period of 0.3 s. Determine the spring stiffness and the viscous damping coefficient.

Solution

ans)

Forces that affect the bucket m2:

F G2FG2…weight

T T…tension force by which the rope affects the bucket

Forces that affect the block m1:

F G1FG1…weight

T T…tension force by which the rope affects the block

T T…tension force by which the rope affects the block

The vector force equation for the bucket:

F G2+T =m2a 2

FG2+T=m2a2

The vector force equation for the block:

G1+T+T=m1a1

FG1+T+T=m1a1

If the bucket goes up a distance s, the block goes down a distance s2s2.

s=12a2t2s=12a2t2

s2=12a1t2s2=12a1t2

We divide the first equation by the second equation, yielding:

2=a2a12=a2a1

a2=2a1(7)(7)a2=2a1

Equation (5) and equation (6) can be written as:

Tm2g=m22a1(8)(8)Tm2g=m22a1

m1g2T=m1a1(9)(9)m1g2T=m1a1

We have two equations (8) and (9) in two variables T and a1. From them we can determine the magnitude of the acceleration a1 and the force T. We multiply equation (8) by 2 and add both equations up:

m1g2m2g=m1a1+4m2a1m1g2m2g=m1a1+4m2a1

(m12m2)g=(m1+4m2)a1(m12m2)g=(m1+4m2)a1

a1=(m12m2)gm1+4m2(10)(10)a1=(m12m2)gm1+4m2

According to equation (7):

a2=2a1=2(m12m2)gm1+4m2=(2m14m2)gm1+4m2(11)(11)a2=2a1=2(m12m2)gm1+4m2=(2m14m2)gm1+4m2

The magnitude of the tension force T is given for example by equation (8):

Tm2g=m22a1Tm2g=m22a1

T=m22a1+m2g=m22(m12m2)gm1+4m2+m2gT=m22a1+m2g=m22(m12m2)gm1+4m2+m2g

T=2(m2m12m22)gm1+4m2+m2g=(2m2m14m22+m2(m1+4m2))gm1+4m2T=2(m2m12m22)gm1+4m2+m2g=(2m2m14m22+m2(m1+4m2))gm1+4m2

T=(2m2m14m22+m2m1+4m22)gm1+4m2=3m2m1gm1+4m2T=(2m2m14m22+m2m1+4m22)gm1+4m2=3m2m1gm1+4m2

T=3m2m1gm1+4m2(12)T=3m2m1gm1+4m2