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,

Irodov Problem 4.6

The equation of motion of the particle is given by,



The time elapsed for 3/8 of the time period of oscillations is given by,




As with problem 4.5, the mean velocity is the displacement per time and so we have,




















The mean of modulus of the velocity is essentially the mean speed of the particle and is given by the the total distance (not displacement) per time. As seen in the figure, in the first 2/8 (1/4) of the time period the particle travels a distance of a to reach the extreme position. After this point it reverses its journey and starts to move towards the center. Over the remaining 1/8 of the period, from (1) we can compute that the particle has reached a point that is distance from the center. This means that the particle has traveled a distance during the remaining 1/8 of the time period. The mean speed of the particle can thus be computed as,

Irodov Problem 4.5

The equation for simple harmonic oscillations of amplitude a and time period T is given by,




The mean velocity over a time interval is the displacement over the time interval over the total time elapsed during the time interval given by,

















From (1) we can see that the oscillating particle will reach half the amplitude (by setting x=a/2) within t1 = T/6 seconds and it will reach the extreme position (setting x = a) in t2=T/4 seconds. This is also shown in the figure. Hence, we have,







Thus we can compute the mean velocities in question as,

Irodov Problem 4.4

As derived in Problem 4.3 Equation (2a) we have the amplitude square of the oscillation as,

Irodov Problem 4.3

Any simple harmonic oscillation can be described by equations (1a) and (1b) given as,

Irodov Problem 4.2

From the problem description we have,








From (1a) it is clear that the amplitude of the oscillations is a/2 and the angular frequency is .

We can now obtain the velocity as a function of x as,

Irodov Problem 4.1

From the problem description we have,