Irodov Problem 4.13

Let the motion of the particle as seen from the frame K' be,



Let the motion of the origin O of the frame K' as seen from frame K be



Clearly the position of the particle as seen from frame K will be,



Since a beat frequency of occurs at f2 = 20Hz and 24Hz we have,



Beats of frequency will thus occur when,

Irodov Problem 4.12

From (1) it is clear that the particle's motion can be decomposed into two cosine waves with and . The beat frequecy then is given by,



Hence the time period of the beats is given by,

Irodov Problem 4.11

Since the particle participates in two different simple harmonic motions its net motion is governed by their summation. The equation of motion of the particle is then given by,










The velocity (1b) is thus the summation of two sine waves shown in the figure. As we trace time starting from t=0 towards we see that both the sine waves are rising. Hence, from t=0 to the velocity v will keep rising. After , one of the sine waves starts to fall while the other continues to rise. Thus, graphically it is clear that the maximum of velocity will lie somewhere at point A that lies in the interval as shown in the figure.


Now the maximum of v will occur when and . In other words,