1 88.] PLANE MOTION. 



lor 



receives a vertical simple harmonic motion from the T bars BG 

 and DH, just as in Fig. 45 (Art. 179). If the free end F of the 

 cord be just kept tight, its vertical displacement will be twice 

 the sum of the vertical displacements of B and D, and as these 

 points have simple harmonic motions, the motion of F will be 

 twice the resultant simple harmonic motion. 



The idea of this mechanism is due to Lord Kelvin. 



188. Exercises. 



( i ) Find the resultant of three simple harmonic motions in the same 

 line, and all of period T= 10 seconds, the amplitudes being 5, 3, and 

 4 cm., and the phase differences 30 and 60, respectively, between the 

 first and second, and the first and third motions. 



(2) Apply the geometrical method of Art. 185 to the problem 

 of Art. 183. 



(3) Find the resultant of two simple harmonic motions in the same 

 line and of equal period when the amplitudes are equal and the phases 

 differ : (a) by an even multiple of TT, () by an odd multiple of TT. 



(4) Resolve x = 10 cos (TT/+ 45) into two components in the same 

 line with a phase difference of 30, one of the components having the 

 epoch o. 



(5) Trace the curves representing the component motions as well as 

 the resultant motion in Ex. (i), taking the time as abscissa and the 

 displacement as ordinate. 



(6) Show that the resultant of n simple harmonic motions of equal 

 period Z'in the same line, viz. : 



^/4- A .- * = 0nCos 

 is the isochronous simple harmonic motion 



= = *! cos * 2 = 



x = a coSi T 



where a* = (2a t cos Ci ) 2 + (%*t sin * 



*_L_ /"?./ cin t.\ 2 



and tan c = 20< sin e^/Stf, cos <. 



