Problems: 15-CQ9, 15-2, 15-14, 15-19, 15-36, 15-42, 15-44, 15-51, 15-60, 15-63 Chapter 15, problem 11 (|| 2): In Fig. 15-29, two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. What is the frequency of oscillation on the frictionless floor?
Write newton’s law, law, and insert the ansatz x x(t ) xm cos(t ) ; this is a differential equation,
F F F 1
2
kx1 kx2 kx k ( x) 2kx m f =
2
=
1
2k
2
m
d2x
mx m( 2 x) m(2 f ) 2 x;
2
dt
1
2(7580 N/m)
2
0.245 kg
(1.1)
39.6 Hz ;
cos(3t 13 ) gives the simple harmonic motion of a body. Chapter 15, problem 9 (|| 14): The function x 6.0 co
At t = 2.0 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f ) period of the motion? (a) motion? (a) Making sure our calculator is in radians mode, we find
b g G J= 3.0 m. H 3K
x = 6.0 cos 3 2.0 +
(b) Differentiating with respect to time and evaluating at t = = 2.0 s, we find dx v= = 3 6. 6.0 sin 3 2. 2.0 + = 49 m / s. dt 3
bgG J Hb g K
(c) Differentiating again, we obtain dv a= = 3 dt
3b 2.0 + J= 2.7 10 b gb6.0 gcosG g H 3K 2
2
2
m/s .
(d) In the second paragraph after Eq. 15-3, the textbook defines the phase of the motion, (t ) t 0 (3 (2.0) 13 ) r ad 20 20 rad
(1.2)
(e) Comparing with Eq. 15-3, we see that = = 3 rad/s. Therefore, f Therefore, f = = /2 /2 = 1.5 Hz. (f) The period is the reciprocal of the frequenc y: T = = 1/ f f 0.67 s.
Chapter 15, problem 18 (|| 19): At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in simple harmonic motion, with a period of 12.5 12 .5 h. How long does it take for the water to fall a distance distance 0.250d from its highest level? level? We We need to translate the problemgivens into the parameters of x (t ) xm cos(t 0 ) , as, xm
xhigh xlow 2
d 0 2
1
1
2
T
d ; 2 f 2
2 12.5hr
0.503
rad
; 0 0; 0; hr
We seek t when when the ocean surface is at x 12 xm 14 d . With x xm cos(t ) , we obtain
(1.3)
1 4
1 rad 1 1 1 1 2.08hr ; (1.4) d cos 0.503t 0 cos(0.503 t) t cos 2 hr 0.503 rad 2 2 hr
d
Chapter 15, problem 35 (|| 36): A 10 g particle undergoes SHM with an amplitude of 2.0 mm, a maximum acceleration of magnitude amax 8.0 103 sm2 , and an unknown phase constant . What are (a) the period of the
motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement? (a) The SHM implies x x(t ) xm cos(t ) , as usual. This means, a(t )
d 2 x(t ) 2
dt
x(t) x 2 xm cos(t 0 ) am cos(t 0); am 2 xm
am xm
8.0 103
m s2
2.0 103 m
2000
rad s
(1.5)
;
(b) max speed can you derive it yourself? dx(t ) rad m xm sin(t 0 ) vm sin(t 0 ) vm xm 2000 2.0 10 3 m 4.0 (1.6) v(t ) dt d s (c) total energy: check this out, 2
2
E K U 12 mv(t ) 2 12 kx(t) 2 12 m xm sin(t 0 ) 12 k xm cos(t 0 )
m xm sin kxm cos m x 1 2
2
2
2
2
1 2
1 2
(c) The total energy is
2
1 2
2
2 m
sin
2
cos
2
1 2
m x 1 2
2 2 m
1 2
2
m x
2 m
mv 1 2
2 m
kx ; 1 2
2 m
(1.7)
kxm2 21 mvm2 0.080 J .
(d) At the maximum displacement, the force acting on the particle is
Fm kxm (4.0104 N/m)(2.0103 m) 80 N mam ; (e) At half of the maximum displacement, x 1.0 mm , and the force is
F kx k 12 xm (4.0104 N/m)(1.0103m) 40 N ; Chapter 15, problem 41 (|| 42): In Fig. 15-40, the pendulum consists of a uniform disk with radius r = 10.0 cm and mass 500 g attached to a uniform rod with length L = 500 mm and mass 270 g. (a) Calculate the rotational inertia of the pendulum about the pivot point. (b) What is the distance between the pivot point and the center of mass of the pendulum? (c) Calculate the period of oscillation. 2 2 (a) using I disk 12 Mr , I rod 13 mL , and the parallel axis theorem for h r L ,
I I disk I rod Mh2 1
1
1 Mr 2 mL2 M ( L r )2 2 3 1
(1.8)
(0.500 kg)(0.100 m) (0.500 kg)(0.500 m 0.100 m) (0.270 kg)(0.500 m) 0.205 kg m ; 2
2
2
3
2
2
(b) We put the origin at the pivot. The center of mass of the rod is r L / 2 (0.500 m) / 2 0. 250 m away, on the same line. The distance from the pivot point to the center of mass of the disk−rod system is xi mi M d m r M ( L r ) m r d xCM xP xCM 0 mi M m M m
0.500 kg 0.500 m + 0.100 m 0.270 kg 0.250 m 0.500 kg 0.270 kg
0.477 m ;
(c) The period of oscillation is T 2
g
2
I ( M m) d
g
2
I ( M m) gd
2
0.205 kg m2
1.50 s . (0.500 kg 0.270 kg)(9.80 m / s 2 )(0.447 m)
Chapter 15, problem 43 (|| 44): (a) If the physical pendulum of Fig. 15-11 and the associated sample problem is inverted and suspended at point P, what is its period of oscillation? (b) Is the period now greater than, less than, or equal to its previous value?
(a) Referring to Sample Problem – “Physical pendulum, period and length,” we see that the distance between P and C is h 23 L 12 L 16 L . The parallel axis theorem (see Eq. 15 – 30) leads to I L2 / 9 2L 1 1 2 1 2 2 2 1.64 s ; (1.9) I = mL + mh = + mL = mL T 2 12 9 mgh gL / 6 3g 12 36 1
2
2
(b) We note that this T is identical to that computed in Sample Problem – “Physical pendulum, period and length.” As far as the characteristics of the periodic motion are concerned, the center of oscillation provides a pivot that is equivalent to that chosen in the Sample Problem (pivot at the edge of the stick).
Chapter 15, problem 48 (|| 51): A rectangular block, with face lengths a = 35 cm and b = 45 cm, is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. Figure 15-43 shows one possible position of the hole, at distance r from the block’s center , along a line connecting the center with a corner. (a) Plot the period of the pendulum versus distance r along that line such that the minimum in the curve is apparent. (b) For what value of r does that minimum occur? There is actually a line of points
ar ound the block’s center for which the period of swinging has the same minimum value. (c) What shape does that line make?
(a) For the “physical pendulum” we have, T 2
I mgh
2
I com mh2 mgh
T
2 g
a 2 b2 12r
r T ( r );
(1.10)
In the figure below, we plot T as a function of r , for a = 0.35 m and b = 0.45 m.
(b) The minimum of T can be located by setting its derivative to zero, dT / dr 0 . This yields
r
a 2 b2 12
(0.35 m)2 (0.45 m)2 12
0.16 m.
(c) The direction from the center does not matter, so the locus of points is a circle around the center, of radius 2 2 1/2 [(a + b )/12] . Chapter 15, problem 58 (|| 60): In a damped oscillator with m = 250 g,k = 85 N/m, and b = 70 g/s, what is the ratio of the amplitude of the damped oscillations to the initial amplitude at the end of 20 cycles?
F m
d 2x dt 2
Fdamp Fspring b
In (1.11), we have
k m
dx dt
kx mx bx kx 0; x(t) xm e
2bm t
cos( t 0 ) (1.11)
4bm . Referring to the numbers in Sample Problem – “Damped harmonic oscillator, 2
2
time to decay, energy,” we have m = 0.25 kg, b = 0.070 kg/s, and T = 0.34 s. Thus, when t = 20T , the damping factor becomes
e bt 2m e
gg g
g 0.39.
0.070 20 0 .34 / 2 0.25
Chapter 15, problem 61 (|| nothing): For Eq. 15-45, suppose the amplitude xm is given by xm
F m 2
2
m (d 2 )2 b2 d 2
where Fm is the (constant) amplitude of the external oscillating force exerted on the spring by the rigid support in Fig. 15-14. At resonance, what are the (a) amplitude and (b) velocity amplitude of the oscillating object? (a) We set = d and find that the given expression reduces to xm = F m/b at resonance. (b) In the discussion immediately after Eq. 15-6, the book introduces the velocity amplitude vm = xm. Thus, at resonance, we have vm = F m/b = F m/b.
