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140 HOMEWORK Chapter 15 (AKA 15A) Q4. Which of the following relationships between the a and the x of a particle involve SHM: (a) a =0.5x, (b) a = 400 x2, (c) a = −20 x, (d) a = −3 x2? ——— (c). The acceleration has to be directly proportional to a negative constant the in SHM. Q7. Figure 15-22 shows the x(t) curves for three experiments involving a particular – box system oscillating in SHM. Rank the curves according to (a) the system’s angular , (b) the spring’s potential at t = 0, (c) the box’s at t = 0, (d) the box’s at t = 0, and (e) the box’s maximum kinetic energy, greatest first. ——— (a) All equal. Same time between crests; plus, it’s the same system.

(b) 3 > 2 = 1. The Ue(0) is proportional to the square of the x value at t = 0. (c) 1 > 2 > 3 = 0. K is proportional to the square of the slope of the x vs t curve at t = 0. (d) Same as (c). The speed is proportional to the square root of K.

(e) 1 > 3 > 2. The maximum K will have the same value as the maximum of Ue, which corresponds with the maximum amplitude. P5. In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic , with frequency 120 Hz. Find (a) the amplitude, (b) the maximum blade speed, and (c) the magnitude of the maximum blade acceleration. ——— Note first that ω =2π f = 754 rad/s. Also the conversion to meters (a) The total excursion is 2 mm, so the amplitude is half that: 1mm= 0.001 m. (b) v = −Aω cos ωt, so vmax = Aω = (1 mm)(754 rad/s) = 0.754 m/s. 2 2 2 2 (c) a = −Aω cos ωt, so amax = Aω = (1 mm)(754 rad/s) = 569 m/s . P10. An oscillating block – spring system takes 0.75 s to begin repeating its motion. Find (a) the period, (b) the frequency in , and (c) the angular frequency in per . ——— (a) 0.75 s. Definition of period is time to repeat. (b) f =1/T =1/(0.75 s) = 1.33 Hz. Hertz (Hz) are cycles per second. (c) ω =2π f =6.28 · 1.33 = 8.38 rad/s. P22. Figure 15-32 shows block 1 of 0.200 kg to the right over a frictionless elevated surface at a speed of 8.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of 0.140 s, and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h =4.90 m. What is the value of d? ——— Recognize a multi-part problem that involves SHM, collisions, and projectile motion. To get d, we need

vf of block 1. To get vf,1, we need the mass of block 2. To get m2, we use SHM, needing ω to get started. ω =2π/T = 44.880 rad/s. 2 2 2 ω = k/m2 ⇒ m2 = k/ω = 1208.5/(44.880 )=0.60000 kg.

Use the stationary-block-2 equation for vf,1:

vf,1 =(m1 − m2)vi,1/(m1 + m2)=(−0.4 kg)(8 m/s)/(0.8 kg) = −4m/s.

Projectile motion: calculate the time to hit th for ∆y = −4.9 m for v0y = 0: 2 2 ∆y =0 − (1/2)gth ⇒ th = p−2∆y/g = p−2 · (−4.9 m)/(9.8m/s )=1.000 s. Projectile motion: vx is constant, so

d = |vx|th =(4m/s)(1 s) = 4.00 m. Final answer to 3 sigdig. P23. A block is on a horizontal surface (a shake table) that is moving back and forth horizontally with of frequency 2.0 Hz. The coefficient of static between block and surface is 0.50. How great can the amplitude of the SHM be if the block is not to slip along the surface? ———

We need SHM and Newton’s Second Law. The that accelerates the block is fs, just as it’s fs that accelerates the crate in the back of the pickup truck. We will also need ω: ω =2π f =6.28(2) = 12.56 rad/s.

FBD for block: mg down, n up, fs to right. The static frictional force varies sinusoidally to right and left; I arbitrarily pick when it’s to the right. ~a to right. +x to right; +y upward. y-eq: n − mg =0 ⇒ n = mg. 2 2 SHM: a = −Aω cos ωt, so amax = Aω is the limiting case for the block. 2 x-eq: fs = ma. Sub for fs = fs,max = µsn = µsmg and a = Aω . 2 µsmg = Amaxω ⇒ 2 2 2 Amax = µsg/ω = (0.5)(9.8m/s )/(12.56 rad/s) =0.0311 m. P30. An oscillating block – spring system has a mechanical energy of 1.00 J, an amplitude of 10.0 cm, and a maximum speed of 1.20 m/s. Find (a) the spring constant, (b) the mass of the block, and (c) the frequency of . ——— 2 2 2 (a) Emec = (1/2)kA , so k =2Emec/A =2 · (1 J)/(0.1 m) = 200 N/m. 2 (b) Use the fact that Kmax = Emec = (1/2)mvmax: 2 2 (1/2)mvmax =1J ⇒ m =2 · (1 J)/(1.2m/s) = 1.39 kg. (c) We get ω from the equation and information, then use that to get f = ω/2π: vmax = Aω =1.20 m/s ⇒ ω = (1.20 m/s)/(0.1m)=12.0 rad/s. f = 12.0/6.28 = 1.91 Hz. P47. In Fig. 15-42, a physical consists of a uniform solid disk (of radius R = 2.35 cm) supported in a vertical plane by a pivot located a distance d =1.75 cm from the center of the disk. The disk is displaced by a small angle and released. What is the period of the resulting simple harmonic motion? ——— Physical pendulum period: using Eq. 15-29. Note that 2 2 2 2 2 I = ICM + mh = (1/2)mR + mh = m(R /2 + h ). T =2πpI/mgh =2πpm(R2/2 + h2)/mgh =2πp(R2/2 + h2)/gh = 6.28p[0.5 · (0.0235 m)2 + (0.0175 m)2]/[9.8m/s2)(0.0175 m)] = 0.366 s P67. In Fig. 15-49, three 10000 kg ore cars are held at rest on a mine railway using a cable that is parallel to the rails, which are inclined at angle θ = 30◦. The cable stretches 15 cm just before the coupling between the two lower cars breaks, detaching the lowest car. Assuming that the cable obeys Hooke’s law, find the (a) frequency and (b) amplitude of the resulting of the remaining two cars. ——— We know the ∆x of the cable, which acts as a spring, but need its tension to get the spring constant k. We then use k to calculate the frequency of oscillation. (a) FBD for the group of 3 cars just before the cable broke: mg down; n away from rails; T up-rail. ~a =0. +x up-rail to minimize trig. x-eq: T − mg sin θ =0 ⇒ T = mg sin θ = (3 × 104 kg)(9.8m/s2)(0.5)=1.47 × 105 N. k = T/∆x = (1.47 × 105 N)/(0.15m)=9.8 × 105 N/m. Get ω, then f. ω = pk/m = p(9.8 × 105 N/m)/(2 × 104 kg) = 7 rad/s f = ω/2π = 1.11 Hz. (b) The equilibrium deflection from neutral with 3 cars was 15 cm, which is the initial deflection with two cars. The equilibrium deflection with 2 cars would be 10 cm. Therefore, the initial displacement from the new equilibrium, which is the amplitude, is 5 cm.