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.

No i do not get answer
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Got answer by energy conservation,hope it helps frndz...

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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
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Get ur doubt resolved

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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 k1xand 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 = 1/2k1x+ 1/2 k2x2
v= x2(k1+k2)/m
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1.2 ​k1​ x^2 + 1/2 ​k2 ​x^2=1/2 ​mv^2
v^2=k1​+k2/m ​​x^2
v = root k1​+k2/m ​​​x

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