If the initial position of a block is x = .16 meters and its initial velocity is
ID: 1279985 • Letter: I
Question
If the initial position of a block is x = .16 meters and its initial velocity is - 5 m/s, and K = 100 N/m and M = 3 kg.
c) Write the general equations for x, v, and a.
d) What is the absolute value of the maximum values of x, v, a, K.E., U?
e) When does the velocity 1st reach its maximum value?
f) How many seconds will it take for the PE of spring to equal three times KE of block?
If the initial position of a block is x = .16 meters and its initial velocity is - 5 m/s, and K = 100 N/m and M = 3 kg. a) What is the angular frequency w frequency f, period T? b) What is the amplitude A, phase angle c) Write the general equations for x, v, and a. d) What is the absolute value of the maximum values of x, v, a, K.E., U? e) When does the velocity 1st reach its maximum value? f) How many seconds will it take for the PE of spring to equal three times KE of block?Explanation / Answer
a)
w = sqrt (k/m)
= sqrt (100/3)
= 5.77 rad/s
f = w/(2*pi)
f = 5.77 / (2*3.14)
f = 0.9189 Hz
T = 1/f
T = 1 / 0.9189
T = 1.088 s
b)
KE at the instant = 1/2*mv^2 = 1/2*3*5^2 = 37.5 J
PE at the instant = 1/2*kx^2 = 1/2*100*0.16^2 = 1.28 J
Total energy = KE + PE = 37.5 + 1.28 = 38.78 J
At extreme position, total energy = 1/2*kA^2
38.78 = 1/2*100*A^2
Amplitude A = 0.8806 m
x = A*sin(wt + phi)
v = dx/dt = A*w*cos(wt + phi)
At t = 0,
x = A*sin phi
v = A*w*cos phi
x/v = (1/w)*tan phi
0.16 / (-5) = (1/5.77) tan phi
phi = 2.959 radians
0.16 = A*sin (2.959)
A = 0.882 m
c)
x = 0.882*sin (5.77*t + 2.959)
v = dx/dt = 0.882*5.77 *cos (5.77*t + 2.959)
v = 5.09*cos(5.77*t + 2.959)
a = dv/dt = -5.09*5.77*sin(5.77*t + 2.959)
a = -29.369 *sin(5.77*t + 2.959)
d)
Max. value of x will be equal to A = 0.882 m
Max. value of v = A*w = 5.09 m/s
Max. value of a = A*w^2 = 29.369 m/s^2
Max. KE = 1/2*m*v_max^2 = 1/2*3*5.09^2 = 38.86 J
Max. PE = 1/2*kA^2 = 1/2*100*0.882^2 = 38.9 J
e)
v = 5.09*cos(5.77*t + 2.959)
When v = 5.09 we get, t = 0.576 second.
f)
KE = 1/2*mv^2
KE = 1/2*3*[5.09*cos(5.77*t + 2.959)]^2
PE = 1/2*kx^2
PE = 1/2*100*[0.882*sin (5.77*t + 2.959)]^2
PE / KE = 3
[1/2*100*[0.882*sin (5.77*t + 2.959)]^2] / [1/2*3*[5.09*cos(5.77*t + 2.959)]^2] = 3
Solving this, t = 0.213 seconds