A uniform plank of mass m, free to move in horizontal direction, is placed on the top of a solid cylinder of mass 2m and radius R. The plank is attached with one end of a massless spring of force constant k and the other end of spring is fixed with a wall as shown in fig. The plank is shiftily displaced towards spring and released. There is no slipping any where then

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Total energy of the system, when compression is spring is x and velocity of plank is v, is $\frac{7}{8} m v^{2} + \frac{1}{2} k x^{2}$

Angular frequency of oscillation is $\sqrt{\frac{4 k}{7 m}}$

Time period of oscillation is $π \sqrt{\frac{7 m}{4 k}}$

Angular frequency oscillation is $\sqrt{\frac{2 k}{7 m}}$

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At the instant when velocity of plank is v velocity of C.M. of cylinder is $\frac{V}{2}$ and angular velocity of cylinder is $ω$
$E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} + \frac{1}{2} 2 m {(\frac{v}{2})}^{2} + \frac{1}{2} I {(\frac{V / 2}{R})}^{2} I = \frac{2 m R^{2}}{R} = m R^{2} = \frac{7}{8} m v^{2} + \frac{1}{2} k x^{2}$
For free oscillation $\frac{d E}{d t} = 0 \Rightarrow a = - (\frac{4 k}{7 m}) x$
$\Rightarrow ω^{2} = \frac{4 k}{7 m}$ and $t = \frac{2 π}{ω} = 2 π \sqrt{\frac{7 m}{4 k}}$