## Saturday, May 29, 2010

### Irodov Problem 4.7

At t =0 the position of the particle is x = a (A in the figures). In other words the particle starts at one of the extreme points of the oscillation. Let us start by considering a time t such that . As shown in Figure 1a, the distance traveled by the particle from A to X is given by,

Now let us consider a time t such that . During this time the value of the x coordinate of the particle is negative since the particle is in the left of the origin O. Hwowever, it has already traveled a distance of a while traveling from A to O. As shown in figure 1b, the distance traveled by the particle from A to X is given by,

Now the situation when the particle returns from the other extreme point A' the situation is identical to when the particle started its journey from A except for the fact that the particle has already traveled as distance of 2a by this time. Hence, we have,

In general, at any given time t, the particle has completed traversing n times the distance between A and A' (2a) where n is given by,

Hence the total distance traversed is given by,

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