5. (a) We find the effective spring constant from the frequency: f_{1}= (k/m_{1})^{1/2}/2p; 10 Hz = [k/(0.60 x 10^{-3}kg)]^{1/2}/2p, which gives k = 2.4 N/m. (b) The new frequency of vibration will be f_{2}= (k/m_{2})^{1/2}/2p = [(2.37 N/m)/(0.40 x 10^{-3}kg)]^{1/2}/2p = 12 Hz.

17. We use a coordinate system with down positive. With x_{0}positive, at the equilibrium position we have SF = -kx_{0}+ mg = 0. If the spring is compressed a distance x (so x is negative) from the equilibrium position, we have SF = - k(x + x_{0}) + mg. When we use the equilibrium condition, we get SF = F = - kx. Note that x is negative, so the restoring force is positive (down). If we stretch the spring a distance x (so x is positive), we still have SF = - k(x + x_{0}) + mg. When we use the equilibrium condition, we get SF = F = - kx. Note that x is positive, so the restoring force is negative (up).

29. We find the spring constant from the compression: k = F/x = (95.0 N)/(0.185 m) = 514 N/m. The kinetic energy will be maximum when the ball leaves the gun, and thus equal to the initial potential energy: 1/2mv^{2}= 1/2kx^{2}; (0.200 kg)v^{2}= (514 N/m)(0.185 m)^{2}, which gives v = 9.37 m/s.

31. We can compare the two maximum potential energies: U_{1}/U_{2}= 1/2k_{1}A_{1}^{2}/1/2k_{2}A_{2}^{2}= (k_{1}/k_{2})(A_{1}/A_{2})^{2}; 10 = 2(A_{1}/A_{2})^{2}, or A_{1}= 2.24A_{2}.

44. We use energy conservation between the release point and the lowest point: K_{i}+ U_{i}= K_{f}+ U_{f}; 0 + mgh = 1/2mv_{0}^{2}+ 0, or v_{0}^{2}= 2gh = 2gL(1 - cos q). When we use a trigonometric identity, we get v_{0}^{2}= 2gL(2 sin^{2}1/2q). For a simple pendulum q is small, so we have sin 1/2q ==> 1/2q. Thus we get v_{0}^{2}= 2gL2 (1/2q)^{2}, or v_{0}= q(gL)^{1/2}.

35. For the collision of the bullet and block momentum is conserved: mv = (m + M)V, so V = mv/(m + M). The kinetic energy of the bullet and block immediately after the collision is stored in the potential energy of the spring when the spring is fully compressed: 1/2(m + M)V^{2}= 1/2kA^{2}; (m + M)[mv/(m + M)]^{2}= kA^{2}; [(0.007870 kg)v]^{2}/(0.007870 kg + 6.023 kg) = (142.7 N/m)(0.09460 m)^{2}, which gives v = 352.6 m/s.

10. The dependence of the frequency on the mass is f = (k/m)^{1/2}/2p. Because the spring constant does not change, we have f_{2}/f = (m/m_{2})^{1/2}; (0.48 Hz)/(0.88 Hz) = [m/(m + 1.25 kg)]^{1/2}, which gives m = 0.53 kg.

16. (a) We find the derivatives of the given expression: x = a sin wt + b cos wt; dx/dt = wa cos wt - wb sin wt; d^{2}x/dt^{2}= - w^{2}a sin wt - w^{2}b cos wt = - w^{2}x. Thus Eq. 14-3 is satisfied if w^{2}= k/m. (b) We use a trigonometric identity to expand Eq. 14-4: x = A cos (wt + f) = A(cos wt cos f - sin wt sin f). If we compare this to the given solution, we see that a = - A sin f, and b = A cos f.

22. The impulse, which acts for a very short time, will change the momentum of the mass, giving it an initial velocity. Because this occurs at the equilibrium position, the velocity will be the maximum velocity v_{0}. Thus we have J = Dp = m Dv = mv_{0}. The angular frequency is w = (k/m)^{1/2}, so v_{0}= Aw = A(k/m)^{1/2}. The SHM starts at the equilibrium position, x = 0 at t = 0, so we use a sine function: x = A sin wt = v_{0}(m/k)^{1/2}sin (k/m)^{1/2}t = (J/m)(m/k)^{1/2}sin (k/m)^{1/2}t = [J/(mk)^{1/2}] sin (k/m)^{1/2}t.

36.(a) The period of the motion is independent of amplitude: T = 2p(m/k)^{1/2}= 2p[(0.650 kg)/(184 N/m)]^{1/2}= 0.373 s. The frequency is f = 1/T = 1/(0.373 s) = 2.68 Hz. (b) Because the mass is struck at the equilibrium position, the initial speed is the maximum speed. We find the amplitude from v_{0}= Aw = 2pfA; 2.26 m/s = 2p(2.68 Hz)A, which gives A = 0.134 m. (c) The maximum acceleration is a_{max}= w^{2}A = (2pf )^{2}A = [2p(2.68 Hz)]^{2}(0.134 m) = 38.0 m/s^{2}. (d) Because the mass starts at the equilibrium position, we have a sine function. If we take the positive x-direction in the direction of the initial velocity, we have x = A sin (wt) = A sin (2pft) = (0.134 m) sin [2p(2.68 Hz)t] = (0.134 m) sin [(16.8 s^{-1})t]. (e) We find the total energy from the maximum kinetic energy: E = K_{max}= 1/2mv_{0}^{2}= 1/2(0.650 kg)(2.26 m/s)^{2}= 1.66 J. (f) We find the kinetic energy from the total energy: E = K + 1/2kx^{2}; 1.66 J = K + 1/2(184 N/m)[(0.40)(0.134 m)]^{2}; which gives K = 1.40 J.

45. This is most easily found from the associated circular motion for SHM. We see from the diagram that, out of the 2p radians for a complete circular motion, the total angle corresponding to the pendulum being between + 1/2A and - 1/2A is 4q. We find q from sin q = 1/2A/A = 1/2, so q = p/6. Thus the fraction of time spent between + 1/2A and - 1/2A is fraction = 4(p/6)/2p = 1/3.

80. The object will leave the surface when the maximum acceleration of the SHM becomes greater than g, so the normal force becomes zero. For the pebble to remain on the board, we have a_{max}= w^{2}A = (2pf)^{2}A < g; [2p(5.0 Hz)]^{2}A < 9.80 m/s^{2}, which gives A < 9.93 x 10^{-3}m = 0.99 cm.