Applied Maths - Mechanics 48

還不明白的薑

2009-08-19 10:31 pm

A particle moves in a plane with an acceleration which is parallel to the axis of y and varies as the distance from the axis of x. Show that

(a) when the acceleration is a repulsion, the equation of its path is

y = Aa^x + Ba^(-x) ,

(b) when the acceleration is attractive, the equation of its path is

y = Acos(ax + B) .

回答 (1)

用戶9112

2009-08-20 3:29 am

✔ 最佳答案

(a) d2x/dt2 = 0 => x = pt + q or t = (x – q)/p where p and q are constants
d2y/dt2 = k2y => y = mekt + ne-kt where k, m and n are constants
y = mek(x – q)/p + ne-k(x – q)/p
y = me-kq/pekx/p + nekq/pe-kx/p
y = me-kq/p(ek/p)x + nekq/p(ek/p)-x
Combining constants A = me-kq/p; B = nekq/p; a = ek/p
y = Aax + Ba-x
(b) d2y/dt2 = - k2y => y = mcos kt + nsin kt where m and n are constants
Let tan s = n/m then sin s = n/√(m2 + n2); cos s = m/√(m2 + n2)
y = √(m2 + n2) (cos kt cos s + sin kt sin s)
y = √(m2 + n2)cos(kt – s)
y = √(m2 + n2)cos[k(x – q)/p – s]
y = √(m2 + n2)cos(kx/p – kq/p – s)
Combining constants A = √(m2 + n2); a = k/p; B = -kq/p – s
y = Acos(ax + B)

👍 0 👎 0