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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Oscillations

Eight blocks of masses m , 2m , 3m ,------,8m are placed on a table as shown in the figure .The table is smooth . The block 'm' is given a velocity $\;v_{0}\;$ which inturn hits '2m' and the process continues and finally the systems sticks to a spring of spring constant $\;K_{1}\;$ . Find the angular frequency (w) of the system . (Consider all collisions to be inelastic e=0 )

$(a)\;\large\frac{1}{12}\;\sqrt{\large\frac{K}{m}}\qquad(b)\;\large\frac{1}{3}\;\sqrt{\large\frac{K}{m}}\qquad(c)\;\large\frac{1}{6}\;\sqrt{\large\frac{K}{m}}\qquad(d)\;None\;of\;these$

1 Answer

Answer : (c) $\;\large\frac{1}{6}\;\sqrt{\large\frac{K}{m}}$
Explanation :
Angular frequency dependent only on the mass and spring constant as $\;w=\sqrt{\large\frac{K}{m}}$
Therefore , The total mass that sticks is
$m+2m+-------+\theta m=36 m \qquad \; $ (inelostic collision)
$w=\sqrt{\large\frac{K}{36m}}=\large\frac{1}{6}\;\sqrt{\large\frac{K}{m}}\;.$
answered Mar 10, 2014 by yamini.v
 

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