2丌 (b) The period Tof the oscillation is T 133.3 (c) The maximum speed of the oscillation is vmax =Ao=0.1x1333=1333 m/s The magnitude of the maximum acceleration is amax=Ao=0.1x1333=1.78 x10 m/s 2. A 5 13kg object moves on a frictionless surface under the influence of a spring with force constant88N/cm. The object is displaced 53. 5cm and given an initial velocity of 11. 2m/s back toward the equilibrium position. Find(a) the frequency of the motion, ( b)the initial potential energy of the system,(c)the initial kinetic energy, and(d) the amplitude of the motion Solution (a)The frequency of the motion is I 2.2l(Hz) 2丌2xVm (b) The initial potential energy of the system is PE=kx=-x988x05352=141.4(J) 2 (c) The initial kinetic energy of the system is KE=-mmv-=x5.13x11 2=28. 730) (D)As the total kinetic energy is E= PE+KE=kA, Thus the amplitude of the motion A=2PE+KE)2×(4144287 059(m) 3. Electrons in an oscilloscope are deflected by two mutually perpendicular electric forces in such a way that at any time t the displacement is given by x=Acos ot and y= Acos(at+o) Describe the path of the electrons and determine its equation when(a),=0,(b)8,=30 and(c),=90° (a)When =0, x=Acos ot, y= Acos t, Thus y=x, it is a straight line (b)When =30(b) The period T of the oscillation is 0.54(s) 133.3 2 2 = = = π ω π T (c) The maximum speed of the oscillation is 0.1 133.3 13.33 m/s vmax = Aω = × = The magnitude of the maximum acceleration is 2 2 3 2 amax = Aω = 0.1×133.3 = 1.78×10 m/s 2. A 5.13kg object moves on a frictionless surface under the influence of a spring with force constant 9.88N/cm. The object is displaced 53.5cm and given an initial velocity of 11.2m/s back toward the equilibrium position. Find (a) the frequency of the motion, (b) the initial potential energy of the system, (c) the initial kinetic energy, and (d) the amplitude of the motion. Solution: (a) The frequency of the motion is 2.21(Hz) 2 1 2 = = = m k π π ω ν (b) The initial potential energy of the system is 988 0.535 141.4(J) 2 1 2 1 2 2 PE = kx = × × = (c) The initial kinetic energy of the system is 5.13 11.2 28.73(J) 2 1 2 1 2 2 KE = mv = × × = (D) As the total kinetic energy is 2 2 1 E = PE + KE = kA , Thus the amplitude of the motion is 0.59(m) 988 2( ) 2 (141.4 28.73) = × + = + = k PE KE A 3. Electrons in an oscilloscope are deflected by two mutually perpendicular electric forces in such a way that at any time t the displacement is given by x = Acosωt and cos( ) y y = A ωt +φ . Describe the path of the electrons and determine its equation when (a) o = 0 φ y , (b) o = 30 φ y , and (c) o = 90 φ y . Solution: (a) When o = 0 φ y , x = Acosωt , y = Acosωt , Thus y=x, it is a straight line. (b) When o = 30 φ y , ⎪ ⎩ ⎪ ⎨ ⎧ = + = ) 6 cos( cos π ω ω y A t x A t