Oscillations · Question 1

The center of a disk of radius r and mass m is attached to a spring of spring constant k, inside a ring of radius R>r as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as T= 2 . The correct expression for is (g is the acceleration due to gravity): [figure]

1. 23 ( gR-r+ km )
2. 2g3(R-r)+ km
3. 16 ( gR-r+ km )
4. 14 ( gR-r+ km )

As shown in the figures, a uniform rod OO' of length l is hinged at the point O and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end (O') of the rod, as shown in Fig. 1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is f1. On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is f2. Ignoring gravity and assuming motion only in the plane of the diagram, the value of f1f2 is: [figure]

1. 2
2. 2
3. 52
4. 25

Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circular hoop of radius R. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ 0 is the permittivity of free space.]

1. q2/(4 0 R3m)
2. q2/(32 0 R3m)
3. q2/(8 0 R3m)
4. q2/(16 0 R3m)

A block of mass 5 kg moves along the x-direction subject to the force F=(-20x+10) N, with the value of x in metre. At time t=0 s, it is at rest at position x=1 m. The position and momentum of the block at t=( /4) s are

1. -0.5 m, 5 kg m/s
2. 0.5 m, 0 kg m/s
3. 0.5 m, -5 kg m/s
4. -1 m, 5 kg m/s

If the collision occurs at time t0 = /(2 ), then the value of 4b2/a2 will be .