Suppose a 32-pound weight is suspended on a spring.The weight is raised foot above the equilibrium position and started off with a downward velocity of foot per second.
Assume that the weight stretches the spring foot from its natural position.Find a formula for the
Position of the weight as a function of time t.

Question

Suppose a 32-pound weight is suspended on a spring.The weight is raised  foot above the equilibrium position and started off with a downward velocity of  foot per second.
Assume that the weight stretches the spring  foot from its natural position.Find a formula for the
Position of the weight as a function of time t.

Quizplus · Accepted Answer

The position of the weight as a function of time can be represented by the following equation:

x(t) = A cos(ωt) + B sin(ωt) + C

where A, B, and C are constants determined by the initial conditions, and ω is the angular frequency of the oscillation.

The weight has been raised to a height of 11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11 above the equilibrium position, so the initial displacement is x(0) = 11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11.

The weight is initially given a downward velocity of 11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11 per second, so the initial velocity is v(0) = -11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11.

Since the weight stretches the spring 11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11 from its natural position, the equilibrium position is shifted downward by 11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11. Thus, we have C = -11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11.

The angular frequency can be calculated using the formula ω = sqrt(k/m), where k is the spring constant and m is the mass of the weight. Since the weight is suspended on the spring, the weight and the spring move together, so we can use their combined mass. The weight has a mass of 32 pounds, which is equivalent to 14.515 kg. The spring constant can be calculated using Hooke's Law: F = -kx, where F is the force exerted by the spring, x is the displacement from the equilibrium position, and the negative sign indicates that the force is opposite to the displacement. When the weight is suspended on the spring, the force exerted by the weight is balanced by the force exerted by the spring, so we have:

mg = kx

where g is the acceleration due to gravity. Substituting the given values, we get:

(14.515 kg)(9.81 m/s^2) = k(11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11)

Solving for k, we get:

k = (14.515 kg)(9.81 m/s^2)/(11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11)
   = 201.482 N/m

Substituting the values into the formula for ω, we get:

ω = sqrt(k/m) 
   = sqrt(201.482 N/m / 14.515 kg)
   = 3.533 rad/s

To determine the constants A and B, we need to use the initial displacement and initial velocity:

x(0) = A cos(0) + B sin(0) - 11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11
     = A - 11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11

v(0) = -ωA sin(0) + ωB cos(0)
     = ωB

Substituting the values, we get:

A - 11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11 = 11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11
ωB = -11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11

Solving the equations simultaneously, we get:

A = 0
B = -3.111

Substituting the values into the formula for x(t), we get:

x(t) = -3.111 sin(3.533t) - 11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11

Therefore, the answer is E. From the given information, we can use the formula for the position of an object undergoing simple harmonic motion:
x(t) = A*cos(wt) + B*sin(wt) + C
where A, B, and C are constants and w is the angular frequency.

First, we need to find the angular frequency:
w = sqrt(k/m)
where k is the spring constant and m is the mass of the weight.
Since the weight stretches the spring 11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11 foot from its natural position, we know that the force exerted by the weight is equal to the spring force:
k*x = mg
where g is the acceleration due to gravity. Solving for k:
k = mg/x = (32 lb)(32.2 ft/s^2)/(11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11 ft) = 98.36 lb/ft
Using the mass of the weight:
m = 32 lb / (32.2 ft/s^2) = 0.993 kg
So,
w = sqrt(98.36 lb/ft / 0.993 kg) = 9.896 rad/s

Next, we need to find the constants A, B, and C.
At t=0, the weight is at its maximum displacement of 11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11 foot above the equilibrium position, and moving downwards with a velocity of 11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11 ft/s. Therefore:
x(0) = 11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11
dx/dt(0) = -11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11
Also, since the motion is initially at its maximum displacement, we know that B=0.

Taking the derivative of x(t), we get:
dx/dt = -Aw*sin(wt) + Bw*cos(wt)
At t=0, we have:
dx/dt(0) = Bw = -11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11
Therefore, B = -11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11 / w
Substituting this into x(0) = A*cos(0) + 0 + C, we get:
A = 11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11 - C
Substituting all of these constants into the formula for x(t), we have:
x(t) = (11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11 - C)*cos(9.896t) - (11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11/9.896)*sin(9.896t) + C

To find the value of C, we use the fact that the weight is initially at rest when it is released:
dx/dt(0) = -Aw*sin(0) + Bw*cos(0) = 0
Therefore, we have:
C = (11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11*9.896/11ebbd4d_3f18_c035_8a0a_930606c843be_TB8202_11)

Simplifying this expression, we get:
C = 9.977 ft
Therefore, the formula for the position of the weight as a function of time is:
x(t) = (11ebbd4d_3f18_c033_8a0a_efaab5f37fa9_TB8202_11 - 9.977)*cos(9.896t) - (11ebbd4d_3f18_c034_8a0a_b55ab6d3ddea_TB8202_11/9.896)*sin(9.896t) + 9.977
which simplifies to:
x(t) = 1.022*cos(9.896t) - 1.109*sin(9.896t) + 9.977

Therefore, the best choice is E.

Suppose a 32-Pound Weight Is Suspended on a Spring