Question

GENERAL HOOKE Somewhere DEEP BELOW THE EARTH's surface, at an UNKNOWN displacement from the Earth's center, a particle of mass m is dangled from a long string, length L; the particle oscillates along a small arc according to the differential equation dt -0 Here, 0 refers to an angular displacement measured from the vertical and t refers to time. The particle's mass is given by m = 3 kg The length of the string is given by L = 2 meters. Whenever the particle arrives at a location of 0 =(1/12) radians from the vertical, the particle has no instantaneous speed. On both sides of the vertical, that is, 0 =(1/12) radians is repeatedly observed to be a “turning point for the particle’s periodic motion.

Please show all steps and provide any explanations through either text or definitions/equations.

Answer 1

L = 2 m The length of the string is so, the (linean) speed of the pantiele at t= is, V = L 0(+-5) 4. 11 LTW 19 LT T TRY LT2 48 2012 48 = 0.411 m/s So, the particles approximate speed al t = T/4 is = 0.411 m/s V The Free Body Diagram is, 114/11 T a mgsino mg mgcoso

in the The porroticle acceleration is a (shown diagram) e is the tension on the string. Applying Mewtons lane get, we T= mg coso And ma = imgsing (3) How; a = – Ld²0 dt 2 And for small angularo displacement, sino N so froom eq (3) we get, - mL d20 Ita mga dzo ata w а L - L where W 2 W = 9 So, the angulars frequency a depends on and length of the string 2 va of w =