Answer to Question #99613 in Classical Mechanics for dani

Question #99613
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.
1
Expert's answer
2019-11-29T09:59:14-0500

For series connection, Equivalent spring constant Keq is given by

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

and for parallel connection

Keq=K1+K2

Since

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

Thus,

"k=\\omega^2m"

For three different cases

K1=ω12m

K2=ω22m

and

Keq=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

ω21222





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