(1) Addition of 65 kg is a force of 637 N. Thus the spring constant
is k = 637 N/0.028 m = 2.28×104 N/m. The frequency of vibration
is given by f = (k/m)1/2/(2p) = (2.28×104/1065)1/2/6.2832 Hz = 0.74 Hz.
(3) It travels 0.25 m each quarter cycle, so the total distance
traveled is 4×0.25 m = 1.0 m.
(9) (d) The position is given by y = A coswt, with w = 6.28×3 s-1 and A = 0.15 m. (a) The velocity is given by
v = -w A sinwt. At the equilibrium point, the velocity
is maximum (sine term equals 1), so |v| = wA = 18.8×0.15 m/s = 2.8 m/s.
(b) v = v0 (1 - x2/A2)1/2 = 2.8 (1 - 0.12/0.152)1/2 = 2.1 m/s
(c) E = m v02/2 = 0.5×2.82/2 = 2.0 J.
(11) Two springs in parallel are equivalent to a single spring having
twice as big a constant. Thus the frequency, being proportional to
the square root of k is higher than the frequency of a single spring
by 21/2 = 1.414; i.e., the frequency is given by
(2 k/m)1/2/(2p)
(15) The work to compress the spring is W = k x2/2 which gives the
spring constant, k = 2 W/x2 = 2×3 J/(0.12 m)2 = 420 N/m.
The force at 0.12 m is F = 420 N/m×0.12 m = 50 N.
Newton's second law gives the mass, m = F/a = 50 N/15 m/s2 = 3.3 kg.
(21) Momentum is conserved. Thus m v = (m+M)V giving v if we can
find V. But V = wA (oscillatory system), where
w = (k/(m+M))1/2 = 100 s-1. Thus V = 22 m/s, giving finally
v = 22.3×0.625/.025 = 560 m/s.
(29) The period is twice the interval between ``tick'' and ``tock'';
i.e., T = 2.0 s = 2p(l/g)1/2, which gives the length as
l = 9.8×22/4/3.14162 = 0.99 m. Stretching lengthens the
period, so the clock will run slow as the result of adding the weight.
(37) v = (B/r)1/2 for water, so use the data on p. 254 (moduli) and p.
276 (density).
(a) For water, v = (2×109/103)1/2 = 1400 m/s.
For the solids, B is replaced by the elastic modulus, giving
(b) For granite, v = (45×109/2700)1/2 = 4100 m/s.
(c) For steel, v = (200×109/7800)1/2 = 5100 m/s.
(39) The speed is given by the square root of the ratio of tension to
mass per unit length. Thus v = (150 N/(0.55 kg/30 m))1/2 = 90 m/s. Thus the time between supports is t = 30/90 s = 0.33 s.
(45) Because of the inverse square law, the intensity at one kilometer
must be larger by 502, giving 2×106×502 W/m2 = 5×109 W/m2. For an area of 10 m2, the energy rate is
5×109 ×10 W = 5×1010 W.
(51) The harmonics are simply integer multiples of the fundamental; i.e,.
2×440 Hz = 880 Hz, 3×440 Hz = 1320 Hz, and 4×440 Hz = 1760 Hz.
(59) Worked in class.
File translated from TEX by TTH, version 1.95.
On 2 Dec 1999, 13:43.