An object of mass 4 grams hanging at the bottom of a spring with a spring constant 5 grams per second square. Denote by y vertical coordinate, positive downwards, and y = 0 is the spring-mass resting position. TI m y(t) y (a) Write the differential equation y" = f(y, y') satisfied by this system. f(x,y) = -5y/4 Note: Write t for t, write y for y(t), and yp for y' (t). (b) Find the mechanical energy E of this system. E(t) = 2((yp))^2+(5/2)y^2 Note: Write t for t, write y for y(t), and yp for y' (t). (c) If the initial position of the object is y(0) = 1 and its initial velocity is y' (0) = -3, find the maximum value of the object velocity, Umax > 0, achieved during its motion. Umax = sqrt(19/2) Σ Σ m Σ

Elements Of Electromagnetics
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An object of mass 4 grams hanging at the bottom of a spring with a spring constant 5 grams per second square. Denote by y vertical coordinate,
positive downwards, and y = 0 is the spring-mass resting position.
Note: Write t for t, write y for y(t), and yp for y' (t).
(b) Find the mechanical energy E of this system.
E(t) = 2((yp))^2+(5/2)y^2
Note: Write t for t, write y for y(t), and yp for y' (t).
(c) If the initial position of the object is y(0) =
achieved during its motion.
Umax
0
(a) Write the differential equation y' = f(y, y') satisfied by this system.
f(x,y) = -5y/4
= sqrt(19/2)
y(t)
m
M
M
Σ
m
1 and its initial velocity is y' (0) = −3, find the maximum value of the object velocity, Umax > 0,
Transcribed Image Text:An object of mass 4 grams hanging at the bottom of a spring with a spring constant 5 grams per second square. Denote by y vertical coordinate, positive downwards, and y = 0 is the spring-mass resting position. Note: Write t for t, write y for y(t), and yp for y' (t). (b) Find the mechanical energy E of this system. E(t) = 2((yp))^2+(5/2)y^2 Note: Write t for t, write y for y(t), and yp for y' (t). (c) If the initial position of the object is y(0) = achieved during its motion. Umax 0 (a) Write the differential equation y' = f(y, y') satisfied by this system. f(x,y) = -5y/4 = sqrt(19/2) y(t) m M M Σ m 1 and its initial velocity is y' (0) = −3, find the maximum value of the object velocity, Umax > 0,
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