Question

consider two spring mass system the horizontal surface is frictionless show that the frequency of horizontal oscillation of the mass m is given by v^2=v1^2+v2^2

when v1 and v2 are the frequencies at which the block would oscillate if connected only to spring one and only spring 2 respectively.

Answer 1

For series connection, Equivalent spring constant K_eq is given by

$1/Keq=\frac{1}{K1}+\frac{1}{K2}$

and for parallel connection

K_eq=K₁+K₂

Since

${\omega}^2=k/m$

Thus,

$k=\omega^2m$

For three different cases

K₁=ω₁²m

K₂=ω₂²m

and

K_eq=ω²m

Thus for series connection of two masses from spring constant equation

1/ω²=1/ω₁²+1/ω₂²

And for parallel connection of two masses from spring constant equation

ω²=ω₁²+ω₂²