Saturday, May 10, 2008

Irodov Solution 1.152

As the force pulls m2, it stretches the spring and thus transmits the force to m1, pulling it too. The center of mass of the system acts like a point particle with mass (m1+m2) that is constantly pulled with a force F and so the center of mass simply accelerates at a constant rate F/(m1+m2) to the left.

With respect to an observer located on the center of mass of the system, he will see that the masses oscillate in simple harmonic motion. He will also feel as if a constant force (similar to force of gravity) of magnitude F is acting. So basically the whole system accelerates to the right while the mass keep oscillating about the center of mass as shown in the figure.

Suppose that the locations of the masses m1 and m2 are given by x1 and x2. There is only one force acting on m1 at any given point of time namely the force due to tension in the spring given by . Similarly on mass m2 there are two forces acting, i) the force F and ii) the tension in the spring . Hence we have,





Here we use the dot notation,



From (1) and (2) we can write,
















At the maximum and minimum stretched positions of the spring and so for this condition in (4) we have two solutions,






The corresponding total lengths of the spring will simply be .

Irodov Problem 1.151

The tension in the string when it is compressed by a length x is given by . Mass m1 is now pressed against the wall which offers a normal reaction of N. to oppose the force with which the spring pushes it into the wall . As soon as mass m2 is released, the spring will push it away from the wall and so m2 will begin to accelerate. As soon as m2 the spring reaches it uncompressed state there is no tension in the string and so the normal reaction offered by the wall m1 on becomes 0. At this m1 point breaks free from the wall.

Since no work is being done by external forces on the system of masses and spring, its energy is conserved. Initially the energy stored in the system is and this energy will be conserved.

As the point when reaches back to the spring's uncompressed position, suppose than mass m2 is moving with a speed v. At this point mass is m1 stationary and has 0 speed. The total kinetic energy is given by . So we have,






The velocity of the center of mass is thus given by