A spherical mass 'm'is dropped vertically off a building starting from rest and 'y=0'.the mass experience a resistive force 'FR=-bv'.let 'r=b/m'.find the position'y(t)'of mass???
To find the position function, y(t), of the mass dropped vertically off a building and experiencing a resistive force, we can use Newton's second law and solve the resulting differential equation.
1. Start with Newton's second law equation:
F_net = ma = mg - FR = m(d^2y/dt^2) - bv
2. Rearrange the equation to isolate the second derivative of y with respect to t:
d^2y/dt^2 = (g - bv/m)
3. Since r = b/m, substitute r into the equation:
d^2y/dt^2 = (g - rv)
4. Now, write the equation as a second-order linear homogeneous ordinary differential equation:
d^2y/dt^2 + rv = g
5. To solve this differential equation, we assume a general solution of the form:
y(t) = Ae^(kt) + Be^(-kt)
6. Take the first and second derivatives of this general solution:
dy/dt = Ak*e^(kt) - Bk*e^(-kt)
d^2y/dt^2 = Ak^2*e^(kt) + Bk^2*e^(-kt)
7. Substitute these derivatives into the differential equation and simplify:
Ak^2*e^(kt) + Bk^2*e^(-kt) + rk(Ae^(kt) + Be^(-kt)) = g
8. Cancel out the common exponential terms and simplify:
Ak^2 + Bk^2 + rkA + rkB = g
9. To satisfy this equation, we need the coefficients of like terms to be equal. Therefore:
k^2 + rk = 0
10. Solve for k:
k^2 = -rk
k = ±i√(r)
11. Plug the k value into the general solution:
y(t) = Ae^(i√(r)t) + Be^(-i√(r)t)
12. Since real solutions are sought, we can rewrite the general solution in a trigonometric form:
y(t) = Ccos(√(r)t) + Dsin(√(r)t)
13. Apply the initial conditions to determine C and D. The problem states that the mass is dropped from rest, so its initial velocity is zero and it starts at position y = 0. Plugging these values into the position function gives:
y(0) = Ccos(0) + Dsin(0)
0 = C(1) + D(0)
C = 0
14. Simplify the position function:
y(t) = Dsin(√(r)t)
And that is the expression for the position function, y(t), of the mass as a function of time, considering the resistive force.