Please solve part (b). A block of mass m1 = 0.1 kg attached to an ideal spring m
ID: 3281289 • Letter: P
Question
Please solve part (b).
A block of mass m1 = 0.1 kg attached to an ideal spring moves on a frictionless surface. Let x(t) be its displacement as a function of time. (a) At time t = 0, the block passes through the point x = 0 moving to the right (i.e., toward positive x). At time t = 0.5 s, the block reaches its maximum excursion of xm = 10 cm (i) What is the period T of oscillation? (Ans: t = 2 s) (ii) What is the spring constant k? (Ans: 0.1 2 N/m) (iii) What is the maximum velocity vm? (Ans: 0.314 m/s) (iv) Write the full expression for x(t) (b) At time ti = 5T/8, a particle of mass ml and moving with velocity u = 3um/V2 to the right collides completely inelastically with the block (i.e., collides and sticks to the block) of the oscillator in (a). (i) What is the position of the block as a function of time after the collision? (Ans: x() =-VG/2)x,n cos( t (t-ti) + ] where = tan-1 V2 (ii) At what time does the block pass the origin again? (Ans: 1.53 s)Explanation / Answer
mass of block = m = 0.1 kg
ideal spring constant = k = 0.1*pi^2 N/m
frictionless floor
let the equation be
x(t) = Asin(wt + phi)
w = sqroot(k/m) = sqroot(0.1/0.1)pi = pi rad/s
now, Vm = pi/10 m/s
hence
Aw = pi/10 = A*pi
A = 0.1 m
at t = 0, x = 0
so, 0 = 0.1sin(phi)
phi = 0 deg
hence the equation is
x(t) = 0.1sin(pi*t)
also, time period = 2*pi/w = 2*pi/pi = 2 s
b. at t1 = 5T/8 = 10/8 s
v = 3vm/sqroot(2) = 3*pi/10*sqroot(2) = 0.666 m/s
collision with inelastic block
new speed = u
speed of the block at t= 10/8 s = v'
v' = 0.1*pi*cos(pi*10/8) = -0.222 m/s to the left
from conservation of momentum
mv' + mv = 2mu
u = (v' + v)/2
hence u = 0.222 m/s to the right
new mass = 2m = 0.2 kg
neew w = sqroot(k/m) = sqroot(0.1*pi^2/0.2) = 2.22441 = sqrt(2)*pi/T rad/s
new equation of motion
x = A'cos(w(t - t1) + phi) .. (1)
also,
new maximum amplitude = A'
from conservation of energy
0.5*k*(A')^2 = 0.5*k*(x(t1))^2 + 0.5(2m)u^2
0.5*0.1*pi^2*(A')^2 = 0.5*0.1*pi^2*(0.1sin(pi*10/8))^2 + 0.5*4*0.222^2
A' =sqroot(3/2)
x = sqroot(3/2)cos(sqrt(2)*pi(t - t1)/T + phi)
to find phi, x = 0.1sin(pi*10/8) for t = 10/8
so, sqroot(3/2)cos( phi) = 0.1sin(pi*10/8)
hence phi = 180 + arctan(sqrt(2))
ii) x = sqroot(3/2)cos(sqrt(2)*pi(t - t1)/T + phi) = 0
t = 1.52 s