A block of mass m is attached to two unstretched springs of spring constants k1 and k2 as shown in figure.The block is displaced towards right through x and is released. Find the speed of the block as it passed through the mean position.

Got answer by energy conservation,hope it helps frndz...

Let body is displaced x towards right. Let velocity of the body be v at its mean position. By law of conversation of energy 1/2mv²-0=1/2 k1x²+1/2 k2x² mv²=x²(k1+k2) V²=x²(k1+k2) /m V=x root k1+k2 /m

Get ur doubt resolved

Fakavalcaj

Always keep in mind "energy can neither be destroyed nor be created"

when we displace the object by 'x' distance

p.e. for A spring 1/2 k₁x² and for B is 1/2k₂x²

there will be no k.e at that point

now at mean position

p.e. =0
but k.e. = 1/2mv²

according to conservation of energy

1/2mv² = 1/2k₁x² + 1/2 k₂x²
v² = x²(k₁+k₂)/m