Simple Harmonic Motion: Solved Problems with Solutions

Posted on Aug 26, 2024 in Physics

3. (a) The amplitude is half the range of the displacement, or x_m = 1.0 mm.

(b) The maximum speed v_m is related to the amplitude x_m by v_m = ωx_m, where ω is the angular frequency. Since ω = 2πf, where f is the frequency,

(c) The maximum acceleration is

6. (a) Since the problem gives the frequency f = 3.00 Hz, we have ω = 2πf = 6π rad/s (understood to be valid to three significant figures). Each spring is considered to support one-fourth of the mass m_car so that Eq. 15-12 leads to

(b) If the new mass being supported by the four springs is m_total = [1450 + 5(73)] kg = 1815 kg, then Eq. 15-12 leads to

9. The magnitude of the maximum acceleration is given by a_m = ω²x_m, where ω is the angular frequency and x_m is the amplitude.

(a) The angular frequency for which the maximum acceleration is g is given by , and the corresponding frequency is given by

(b) For frequencies greater than 498 Hz, the acceleration exceeds g for some part of the motion.

13. When displaced from equilibrium, the net force exerted by the springs is –2kx acting in a direction so as to return the block to its equilibrium position (x = 0). Since the acceleration , Newton’s second law yields

Substituting x = x_m cos(ωt + φ) and simplifying, we find

where ω is in radians per unit time. Since there are 2π radians in a cycle, and frequency f measures cycles per second, we obtain

16. From the highest level to the lowest level is twice the amplitude x_m of the motion. The period is related to the angular frequency by Eq. 15-5. Thus, and ω = 0.503 rad/h. The phase constant φ in Eq. 15-3 is zero since we start our clock when x_o = x_m (at the highest point). We solve for t when x is one-fourth of the total distance from the highest to the lowest level, or (which is the same) half the distance from the highest level to the middle level (where we locate the origin of coordinates). Thus, we seek t when the ocean surface is at . With , we obtain

which has t = 2.08 h as the smallest positive root. The calculator is in radians mode during this calculation.

20. We note that the ratio of Eq. 15-6 and Eq. 15-3 is v/x = –ωtan(ωt + φ) where ω = 1.20 rad/s in this problem. Evaluating this at t = 0 and using the values from the graphs shown in the problem, we find

φ = tan^-1(–v_o/x_oω) = tan^-1(+4.00/(2 1.20)) = 1.03 rad (or –5.25 rad).

One can check that the other “root” (4.17 rad) is unacceptable since it would give the wrong signs for the individual values of v_o and x_o.

24. To be on the verge of slipping means that the force exerted on the smaller block (at the point of maximum acceleration) is f_max = μ_smg. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is a_m = ω²x_m, where is the angular frequency (from Eq. 15-12). Therefore, using Newton’s second law, we have

which leads to

25. (a) We interpret the problem as asking for the equilibrium position; that is, the block is gently lowered until forces balance (as opposed to being suddenly released and allowed to oscillate). If the amount the spring is stretched is x, then we examine force-components along the incline surface and find

at equilibrium. The calculator is in degrees mode in the above calculation. The distance from the top of the incline is therefore (0.450 + 0.75) m = 1.20 m.

(b) Just as with a vertical spring, the effect of gravity (or one of its components) is simply to shift the equilibrium position; it does not change the characteristics (such as the period) of simple harmonic motion. Thus, Eq. 15-13 applies, and we obtain

30. The total mechanical energy is equal to the (maximum) kinetic energy as it passes through the equilibrium position (x = 0):

mv² = (2.0 kg)(0.85 m/s)² = 1.445 J.

Looking at the graph in the problem, we see that U(x=10)=0.5 J. Since the potential function has the form , the constant is . Thus, U(x) = 1.445 J when x = ± 12 cm.

(a) Thus, the mass does turn back before reaching x = 15 cm.

(b) It turns back at x = 12 cm.

35. The problem consists of two distinct parts: the completely inelastic collision (which is assumed to occur instantaneously, the bullet embedding itself in the block before the block moves through a significant distance) followed by simple harmonic motion (of mass m + M attached to a spring of spring constant k).

(a) Momentum conservation readily yields v’ = mv/(m + M). With m = 9.5 g, M = 5.4 kg, and v = 630 m/s, we obtain

(b) Since v’ occurs at the equilibrium position, then v’ = v_m for the simple harmonic motion. The relation v_m = ωx_m can be used to solve for x_m, or we can pursue the alternate (though related) approach of energy conservation. Here we choose the latter:

which simplifies to

43. (a) A uniform disk pivoted at its center has a rotational inertia of , where M is its mass and r is its radius. The disk of this problem rotates about a point that is displaced from its center by r + L, where L is the length of the rod, so, according to the parallel-axis theorem, its rotational inertia is . The rod is pivoted at one end and has a rotational inertia of mL²/3, where m is its mass. The total rotational inertia of the disk and rod is

(b) We put the origin at the pivot. The center of mass of the disk is

away, and the center of mass of the rod is away, on the same line. The distance from the pivot point to the center of mass of the disk-rod system is

(c) The period of oscillation is

45. We use Eq. 15-29 and the parallel-axis theorem I = I_cm + mh² where h = d. For a solid disk of mass m, the rotational inertia about its center of mass is I_cm = mR²/2. Therefore,

49. This is similar to the situation treated in Sample Problem 15-5, except that O is no longer at the end of the stick. Referring to the center of mass as C (assumed to be the geometric center of the stick), we see that the distance between O and C is h = x. The parallel axis theorem (see Eq. 15-30) leads to

Eq. 15-29 gives

(a) Minimizing T by graphing (or special calculator functions) is straightforward, but the standard calculus method (setting the derivative equal to zero and solving) is somewhat awkward. We pursue the calculus method but choose to work with 12gT²/2π instead of T (it should be clear that 12gT²/2π is a minimum whenever T is a minimum). The result is

which yields as the value of x, which should produce the smallest possible value of T.

(b) With L = 1.85 m and x = 0.53 m, we obtain T = 2.1 s from the expression derived in part (a).

51. If the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod and if the torque tends to pull the rod toward its equilibrium orientation, then the rod will oscillate in simple harmonic motion. If τ = –Cθ, where τ is the torque, θ is the angle of rotation, and C is a constant of proportionality, then the angular frequency of oscillation is and the period is

,

where I is the rotational inertia of the rod. The plan is to find the torque as a function of θ and identify the constant C in terms of given quantities. This immediately gives the period in terms of given quantities. Let be the distance from the pivot point to the wall. This is also the equilibrium length of the spring. Suppose the rod turns through the angle θ, with the left end moving away from the wall. This end is now (L/2) sin θ further from the wall and has moved a distance (L/2)(1 – cos θ) to the right. The length of the spring is now

.

If the angle θ is small, we may approximate cos θ with 1 and sin θ with θ in radians. Then the length of the spring is given by and its elongation is Δx = Lθ/2. The force it exerts on the rod has magnitude F = kΔx = kLθ/2. Since θ is small, we may approximate the torque exerted by the spring on the rod by τ = –FL/2, where the pivot point was taken as the origin. Thus τ = –(kL²/4)θ. The constant of proportionality C that relates the torque and angle of rotation is C = kL²/4. The rotational inertia for a rod pivoted at its center is I = mL²/12, where m is its mass. See Table 10-2. Thus, the period of oscillation is

With m = 0.600 kg and k = 1850 N/m, we obtain T = 0.0653 s.

56. The table of moments of inertia in Chapter 11, plus the parallel axis theorem found in that chapter, leads to

I_P = MR² + Mh² = (2.5 kg)(0.21 m)² + (2.5 kg)(0.97 m)² = 2.41 kg⋅m²

where P is the hinge pin shown in the figure (the point of support for the physical pendulum), which is a distance h = 0.21 m + 0.76 m away from the center of the disk.

(a) Without the torsion spring connected, the period is

T = 2π√(I_P/Mgh) = 2.00 s.

(b) Now we have two “restoring torques” acting in tandem to pull the pendulum back to the vertical position when it is displaced. The magnitude of the torque-sum is (Mgh + k)θ = I_Pα, where the small angle approximation (sinθ ≈ θ in radians) and Newton’s second law (for rotational dynamics) have been used. Making the appropriate adjustment to the period formula, we have

T’ = 2π√(I_P/(Mgh + k)).

The problem statement requires T = T’ + 0.50 s. Thus, T’ = (2.00 – 0.50) s = 1.50 s. Consequently,

k = I_P(4π²/T’²) – Mgh = 18.5 N⋅m/rad.

60. (a) From Hooke’s law, we have

(b) The amplitude decreasing by 50% during one period of the motion implies

.

Since the problem asks us to estimate, we let . That is, we let

so that T ≈ 0.63 s. Taking the (natural) log of both sides of the above equation and rearranging, we find

Note: If one worries about the ω’ ≈ ω approximation, it is quite possible (though messy) to use Eq. 15-43 in its full form and solve for b. The result would be (quoting more figures than are significant)

which is in good agreement with the value gotten “the easy way” above.