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A block of mass M is attached to two unstretched springs of spring constant K 1 K1 and K 2 K2 respectively . the block is displaced to right through a distance x and is then released . find the speed of the block as it passes through the mean position, as it was in starting (whatever the velocity will be square that and then make it divide by the displacement ) ! .

[√k1k2/M].x^2 [(K1+k2)/M].x [K1k2/M(k1+k2)].x None

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2 solutions

Mridul Chaturvedi
Jan 12, 2016

Isn't omega = whole root of ( net Spring constant / mass )

In this case the net Spring c. = k1+k2. And mass given to be M

Therefore omega should be = whole root( k1+k2/M)

And the ans. Should be =whole root(k1+k2/M) *x

But the correct option in your question is given to be (k1+k2/M)x

Pls correct the mistake or make the ans as 'none'

Yeh thanks for the correction. I forgot to use root in the first option !

Ishaan Kaul - 5 years, 5 months ago
Ishaan Kaul
Jan 8, 2016

We know that when it passes through the mean position it will have max. Velocity. V=omega.amplitude Substitute the values and u will get the answer!

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