Mechanics. Assignment 15 Solutions

220A Solutions Assignment 15 14-4. The general description of a mass-spring system is x=Acos(ωt + φ) . (1) Note that eq 1 is one of a handful of equations that one needs to know by heart. Anything that oscillates according to simple harmonic motion follows the same equation. Of course, dx v = dt = -ωAsin(ωt +φ) (2) All that is needed is to evaluate the constants, which we do by applying initial conditions as follows: at t = 0, x = 8.8 cm = Acos(ω·0 + φ) (3) v = 0 = -ω(8.8 cm)sin(ω0 +φ). (4) and To get the angular frequency, ω, we need to use the information that T = 0.75 s. Either look this up (middle of p. 366), or even better reason that when t increases from 0 to T, then the cosine returns to its original value so ωT = 2π. Thus, ω = 2π/T = 2π/0.75 s = 8.38 rad/s. (5) Equations 2 and 3 become 8.8 cm = Acos( φ) (3') and 0 = -8.38 rad/sA(8.8 cm)sin(φ). (4') We may solve the two equations 3' and 4' any way we want. An easy solution is to note that eq 4' requires that φ = 0. Then, from 3', A = 8.8 cm, and x=8.8cmcos(8.38 rad/s·t), (1) from which we may calculate the position or the velocity (from the time derivative) at any time we wish. In particular, at t = 1.8 s x=8.8cm·cos(8.38 rad/s·1.8 s) = - 7.14 cm In this last computation, you have a choice of setting your calculator on "radians" or converting radians to degrees by multiplying the argument of the cosine by unity = 180 °/π radians. I prefer the latter. 4-10. The frequency of a mass-spring system is given by k/m f=√  (1) Therefore, 0.88 Hz = √ k/m  (2) and 0.48 Hz = √ k/(m + 1.25 kg)  (3) Solve for m. One way to do this is to multiply eq 3 by 0.88/0.48 and then set eq 2 = eq 3 k/m = 0.88/0.48 √ k/(m + 1.25 kg) , so √   k/m = (0.88/0.48)2 k/(m + 125 kg). Canceling k leads to m = (m + 1.25 kg)/(0.88/0.48)2 , so 0.702 m = 0.372 kg m = 0.530 kg. 14-13. Since x = 3.8m cos( 7πt/4 + π/6) is a particular case of x=Acos(ωt + φ), (1) we can see that ω = 7π/4 rad/s and φ = π/6 rad. The period T = 2π/ω, so T = 2π/(7π/4 rad/s) = 1.14 s and f = 1/T = 0.875 Hz. The position at t = 0 is x = 3.8m cos( 7π0/4 + π/6) = 3.8m cos(π/6) = 3.29 m dx The velocity v = dt = -(7π/4 rad/s) (3.8 m)sin( 7πt/4 + π/6) which at t = 0 is v = -(7π/4 rad/s) (3.8 m)sin(π/6) = -10.4 m/s, and at t = 2 s v = -(7π/4 rad/s) (3.8 m)sin( 7π·2s/4 + π/6) = 18.2 m. dv The acceleration a = dt = - (7π/4 rad/s)2 (3.8 m)cos( 7πt/4 + π/6) which at t = 2s is equal to a = - (7π/4 rad/s)2 (3.8 m)cos( 7π·2s/4 + π/6) = - 57.4 m/s2. 14-14. The tuning fork moves according to x=Acos(ωt + φ) . (1) with ω = 2πf = 2π·264 Hz = 1659 rad/s and A = 1.5 x 10-3 m. Since v = dx = -ωAsin(ωt +φ), the maximum value of v is when the sine is ±1 so the dt maximum speed is ωA = 1659 rad/s ·1.5 x 10-3 m = 2.49 m/s. Similarly, taking another derivative, a = - ω2Acos(ωt + φ), so the maximum value of a is ω2A = (1659 rad/s)2·1.5 x 10-3 m = 4.13 x 103 m/s2. 14-26. The total energy (kinetic + potential) of the SHO is 1 1 E = 2 mv2 + 2 kx2 1 1 = 2 m(ωA) 2sin2(ωt + φ) + 2 kA2cos2(ωt + φ). Since E = constant, we can evaluate E anywhere, for example, when x = A; 1 i.e., when v = 0. Thus E = 2 kA2. The potential energy is one-half E (and so is the kinetic energy) when 1 2 1 1 2 kx = 2 2 · 2 kA , x = ± A/√ 2 1 1 When x = A/2, the potential energy = 2 kx2 = kA2/4 = E/4. So the 2 potential energy is 1/4 the total and thus the kinetic energy is 3/4. 14-27. The mass moves according to x = 0.650 m cos (8.40 rad/s·t). The amplitude is the maximum value of x which occurs when the cosine is unity, so A = 0.650m. The value of ω is 8.40 rad/s from which f = ω/2π 1 1 = 1.34 Hz. The total energy E = 2 m(ωA)2 = 2 2 kg(8.40 rad/s·0.650m)2 1 1 = 29.8 J. The fraction of the total energy that is potential is 2 kx2/ 2 kA2. At x= 0.260 m, this fraction is (0.260m/0.650 m)2 = 0.16. Thus the potential energy is 0.16·29.8 J = 4.77 J and the kinetic energy is 0.84·29.8 J = 20.8 J. 14-68. To proceed, we must assume that energy is conserved (which is only approximate). Thus the potential energy of the jumper is mgh, were h = 21.1 m is the distance above the final position when the net is 1 stretched. At this point the potential energy of the net is 2 kx2 where x = 1.1 m is the amount stretched; x is positive downward. Thus 1 mgh = 2 kx2 1 52 kg·9.8 m/s2·21.1 m = 2 k(1.1 m)2 from which we calculate k = 1.79 x 104 N/m. If the person were lying on the net, her weight mg would equal the upward force due to the net kx, where x is the amount stretched. Therefore, x = mg/k = 52 kg·9.8 m/s2/1.79 x 104 N/m = 2.86 x 10-2 m. If the person were to jump from 35 m, then h = 35m + x and 1 mg(35m + x) = 2 kx2 1 52 kg·9.8 m/s2·(35m + x) = 2 (1.79 x 104 N/m)x2 0 = 17.6 m-1·x2 - (35m + x) . Thus x= +1 ± √ (-1) 2 -4·17.6 m -1 ·(- 35m) = 2.84 x 10-2 m ± 1.41 m. 2·17.6 m -1 So x = - 1.38 m or 1.44 m. Since x is positive downward, x = 1.44 m. There is a reward of one ice-cream cone for anyone who can come tell me the meaning of the x = - 1.38 m result. 14-69. The period of oscillation is approximately T = 2π\R(,L/g) = 0.63 m/9.8 m/s2 = 1.59 s, so f = 1/T = 0.628 Hz. The total energy 2π√  can be calculated at the highest point E = mgh (since there is no kinetic energy), where h is the height above the lowest point. From the diagram, L = h + Lcosθ, so h = L[1 - cos(15°)] = 0.63 m·(3.41 x 10-2) = 2.14 x 10-2 m. Lcos θ θ h The energy at the beginning mgh is equal to the energy when the bob passes the lowest point 1 mgh = 2 mv2, so v=±√ 2gh = ± 0.648 m/s.  Evaluate E anywhere you want since it is constant. For example at the highest point, since the kinetic energy is zero, E = mgh = 0.365 kg·9.8 m/s2·2.14 x 10-2 m = 7.65 x 10-2 J.