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A mass m can rotate with angular velocity w1 on a smooth horizontal table being attached to a spring whose other end having passes through a hole in the table, supports another mass m.This other mass turns around as a conical pendulum with angular velocity w2. if l1 and l2 are the lengths of the string on and below table then
a)l1:l2=w1^2:w2^2
b)l1:l2=w2^2:w1^2
c)l1:l2=w1:w2
d0 none
Hi Kritika,
Let "x" be the elongation in the spring.
So for the mass above the table
Kx = ml1w12
And for the mass below the table, let "Q" be the angle made by the spring with the vertical.
So kxsinQ = mrw22.
From the right triangle l2sinQ = r
So kxsinQ = m(l2sinQ)w22.
Hence l1:l2 = w22:w12.
Optoin (B)
Hope that helps.
Best Regards,
Ashwin (IIT Madras).
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