Oscillations

 An object attached to a spring exhibits simple harmonic motion with an amplitude of 4.0 cm. When the object is 2.0 cm from the equilibrium position, what percentage of its total mechanical energy is in the form of potential energy? ◦ (a) One-quarter. ◦ (b) One-third. ◦ (c) One-half. ◦ (d) Two-thirds. ◦ (e) Three-quarters.

 Express the ratio of the potential energy of the object when it is 2.0 cm from the equilibrium position to its total energy:  U(x) =.5Kx^2 = x^2= 2^2 = 4  E total = 5KA^2= A^2 = 4^2 = 16  4/16 = ¼ the total energy

 Two systems each consist of a spring with one end attached to a block and the other end attached to a wall. The identical springs are horizontal, and the blocks are supported from below by a frictionless horizontal table. The blocks are oscillating in simple harmonic motions with equal amplitudes. However, the mass of block A is four times as large as the mass of block B. How do their maximum speeds compare? ◦ (a) Va max = Vb max ◦ (b) Va max = 2Vb max ◦ (c) Va max =.5 Vb max ◦ (d) This comparison cannot be done by using the data given.

 Va max = (K / m * A) ^1/2  Vb max = (K / 4m * A) ^1/2 ◦ Since A and K are constant, can make a portion where ◦ Va max/ Vb max = (m / 4 M) ^1/2 ◦ Vb = ½

 Two simple pendulums are related as follows. Pendulum A has a length LA and a bob of mass mA; pendulum B has a length LB and a bob of mass mB. If the frequency of A is one-third that of B, then ◦ (a) LA = 3LB and mA = 3mB, ◦ (b) LA = 9LB and mA = mB, ◦ (c) LA = 9LB, regardless of the ratio mA/mB, ◦ (d) A = (3LB)^1/2, regardless of the ratio mA/mB.

 f= 1/ 2pie* (g/L)^.5 Or L = g / 4 pie^2 f^2  La = g / 4 pie^2 f^2  Lb = g / 4 pie^2 f^2 ◦ G is constant, and pie ^2 is constant ◦ Just need to compare La, Lb, fa and fb ◦ So La/Lb = f^2 / (1/3)fb^2 = 9 ◦ Answer C

 A mass of 2 kg is attached to a spring with constant 18 N/m. It is then displaced to the point x = 2. How much time does it take for the block to travel to the point x = 1 ?

 X = Acos (wt) ◦ W = 2 pie / T = (k/m)^1/2 ◦ w= (18/2)^1/2 = 3 ◦ 1 = 2 cos ( 3 t ) ◦ T =.35 seconds

 A mass of 2 kg oscillating on a spring with constant 4 N/m passes through its equilibrium point with a velocity of 8 m/s. What is the energy of the system at this point? From your answer derive the maximum displacement, x m of the mass.

 K =.5 mv^2 =.5(2)(8)^2 = 64 joules  U =.5 kx^2  Since energy is conserved: ◦ K =.5 mv^2 = U =.5 kx^2 = 64 ◦ X m = (x *2 / k) ^1/2 = 32^1/2 = 5.65 almost

 One end of alight spring with a spring constant 10 N/m is attached to a vertical support, while a mass is attached to the other end. The mass is pulled down and released, and exhibits simple harmonic motion with a period of.2 pie. The mass is ◦.1 kg ◦.25 kg ◦.4 kg ◦.04 kg ◦.025 kg

 T= 2 pie (m/k)^1/2  M = kT^2 / 4 pie^2 ◦ = (10)(.2 pie)^2 / 4 pie ^2  M=.1 kg