-Two blocks, of masses m 1 and m 2, are placed on a horizontal surface and attac

Question

-Two blocks, of masses m1 and m2, are placed on a horizontal surface and attached to opposite ends of a spring as in (Figure 1) . The blocks are then pushed toward each other to compress the spring. The spring constant for the spring is k. Ignore friction and the mass of the spring.

Describe the motion of the center of mass of the system when the blocks are released. (move to the left, move to the right, will not move?)

Derive an expression for the angular frequency ? of the motion. Express your answer in terms of k, m1 and m2.

What is ? in the limit m2?m1? Express your answer in terms of k, m1 and m2.

Explanation / Answer

mass of the two blocks m1 and m2

spring constant id k

a)

when the two blocks are released,

center of mass will not move, it remains at rest,

because, No external force is acting on the system

b)

let

x1 is compression due to m1

x2 is compression due to m2

net compression xo=x1+x2 ---(1)

since external force is zero,

m1*x1=m2*x2 --(2)

from (1) and (2)

x1=(m2/(m1+m2))*x0

and

x2=(m1/(m1+m2))*x0

tension force in the spring is,

F=k*xo

====>

F1=k*((m1+m2)/m2)*x1 or F2=k*((m1+m2)/m1)*x2 -----(3)

acceleration , a=F1/m1

a=k*((m1+m2)/(m1*m2)*x1

but,

a=w^2*x1 ( here, w is angular frequency)

====>

(w^2)=k*((m1+m2)/(m1*m2)

W^2=k*((m1+m2)/(m1*m2)

W=sqrt(K(m1+m2)/(m1*m2))

c)

W=sqrt(K(m1+m2)/(m1*m2))

if m2>>m1

====> W=sqrt(K/m1)

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