x°
4.69. (a) ao — ,a = — n/2, when xo >
(b) ao =
TE/2, when xo <0;
= I xo^ ±(13/co)^2 , a = arctan (-13/4 with —3-[/2 <a < 0, if
xo > 0 and n/2 < a< a, if xo <O.
4.70. 13= co -Viz— 1 --- 5 s-1.
4.71. (a) v (t) = ao /cat P 2 e-Ot; (b) v (t) = I xo IV" 1 ± ((3/co) 2 e-Pt.
4.72. The answer depends on what is meant by the given ques-
tion. The first oscillation attenuates faster in time. But if one takes
the natural time scale, the period T, for each oscillation, the second
oscillation attenuates faster during that period.
4.73. 'A = nX0/]I1 + (1— nz) (X0/ 2 n) 2 = 3.3, n' ---1/1 (2n/ke) 2 =
= 4.3 times.
4.74. T = 1/- (4a 2 + X 2 ) AxIg = 0.70 s.
4.75. Q= an/ln = 5 .10 2.
4.76. s 1(1 +e-2 42 )/(1 —e-k/ 2 ) = 2 m.
4g 24.77. (^1) /2 V
T
1 — L 3 10 2.
4.78. T = V 3 / 2 (4n 2 H-X 2 filg ) = 0.9 s.
4.79. co—A/T, 71 ;2c (^2) ( 1111 : 2 ) 2
4.80. 1 1 = AhllaR 4 T.
4.81. T = 2R11a 4 B 2.
4.82. (a) T = 2n -11m/x = 0.28 s; (b) n = (xo — 6)/4,a, ---- 3.5 oscilla-
tions, here = kmg/x.
F
4.83. x (^) ,,,2 ol coo (cos coot — cos cot).
4.84. The motion equations and their solutions:
t < x o4x Flm, x = (1 — cos coot) Flk,
..
t v, X ± CO:x = 0, x = a cos [coo (t — T)
where co( 2 , = klm, a and a are arbitrary constants. From the conti-
nuity of x and x at the moment t = x we find the sought amplitude:
a = (2F1k) I sin (coot/2)1.a (^) 4amg
/1 —(k/23- ) 2 g
4.85. cores= V 14-01,/2n) 2 A/ ' res
A,F0A/
4.86. (Ores= ( 0 4 ± COD 12 = 5.1 .10z s-1.
(1+
4 n 2
262
4.87. (a) coo = licol o.) 2 ; (b)^ =- 0)2 -(011 /2V-5,^ Co =
="iwz—(0)^2 —(^0 i)^2 /^12.
4.88. ri = (1 + 2,., 2 /431 2 ) n/A. = 2.1.
4.89. A = naFo sin cp.
327