Question

Two identical strings of mass 5.5 g and length 75 cm are attached to the ends of a uniform rod weighing 165 N, suspending it horizontally from the ceiling. An additional 185-N weight is attached a quarter of the way between the ends of the rod. If fundamental frequencies of standing waves are set in motion in the two strings, what is the frequency of the beats they will produce? How will the situation change if the additional weight is repositioned exactly at the center of the rod?

Answer 1

solution LLLLLLLLLLLLL T+T5 = 18+16 . = 350 -Whit KT2 Also the sum altorques CoQ, of all forces about " the centre el gravity ... 18sn de w=165 H must be equal to bero. so Ti xb - Taxha -185x4 20 = 2T1 - 979 = 185 OR (2) T- Tg = 9215 By solving equations (1) and 2 weger 7= 221 25 M TO = 128.75 N 55x10-3 M= Linear mats density = 75X1072 = 7.333 X10 so frequency of Fundamental standing waves = f = 2x0.75 L = 115.8 HZ and f2 = 22075 T = 28.34 to a frequencey of beats = | 27,467 If adolitional weight is repositioned at the center, then tension in both string will be equal, and so frequencies fr and fz are also equal. sonumber of beats = [0]