A mass weighing 20 pounds stretches a spring 6 inches The ma

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 8 inches below the equilibrium position. Find the position x of the mass at the times t = pi/12, pi/8, pi/6, pi/4, and 9 pi t/32 s. (Use g = 32 ft/s^2 for the acceleration due to gravity.) What is the velocity of the mass when t = 3 pi/16 s? In which direction is the mass heading at this instant? downward upward ? At what times does the mass pass through the equilibrium position?

Solution

Given that

mass of spring is 20 pounds

also it streches 6 inches when held vertical.

so we have spring constant k = mg/d = 20g/0.5,

angular velocity w = sqrt(k/m) = sqrt (g/d) = sqrt (32/0.5) = 8 rad/sec

we know x = a sin(wt+ phi)

here x is distance of mass from equilibrium position. upward is positive and doward is negetive.

so at t = 0; x = -8 inches

x = -8= a sin(phi)

here a = 8, and phi = 3 pi/2;

we can write

x = 8 sin (8t - pi/2)

1) now t = pi/12

we get x = 8 sin(2pi/3 - pi/2) = 8 sin( pi/6) = 4 inches; above equilibrium position

2) now t = pi/8

we get x = 8 sin(pi - pi/2) = 8 sin( pi/2) = 8 inches; above equilibrium position

3) now t = pi/6

we get x = 8 sin(4 pi/3 - pi/2) = 8 sin( 5 pi/6) = 4 inches; above equilibrium position

4) now t = pi/4

we get x = 8 sin(2pi - pi/2) = 8 sin(- pi/2) = -8 inches; below equilibrium position

5) now t = 9pi/32

we get x = 8 sin(9pi/4 - pi/2) = 8 sin( -pi/4) = -4sqrt (2) inches; below equilibrium position

6) velocity of mass is v = 64 cos(8t - pi/2)

at t= 3 pi/16

v = 64 cos(pi) = -64 ft/sec; heading downwards.

7) when 8t - pi/2 = n pi

it passes through equilibrium position.

t = (n+2)/16 pi