Assuming the spring obeys Hooke's law:
F=kxF=kx
Where
FF=force
kk=spring constant
xx= extension
The question implies that alphaα and betaβ is the total length of the spring, rather than the extension. If we define the length of the spring without any weights on it as yy, then
We know that
4=k(alpha-y)4=k(α−y) .......equation (1)
and
5=k(beta-y)5=k(β−y) .......equation (2)
If we divide the two equations:
5/4=(beta-y)/(alpha-y)54=β−yα−y
We can expand and solve for y
5(alpha-y)=4(beta-y)5(α−y)=4(β−y)
5alpha-5y=4beta-4y5α−5y=4β−4y
5alpha-4beta=y5α−4β=y ..........equation (3)
For the 9N weight, of total length, say zz:
9=k(z-y)9=k(z−y)
If we divide this by equation (2)
9/5=(z-y)/(beta-y)95=z−yβ−y
9beta-9y=5z-5y9β−9y=5z−5y
9beta-4y=5z9β−4y=5z
Substitute in for y from equation (3):
9beta-4(5alpha-4beta)=5z9β−4(5α−4β)=5z
9beta-20alpha+16beta=5z9β−20α+16β=5z
25beta-20alpha=5z25β−20α=5z
z=5beta-4alphaz=5β−4α
