Question

A weight stretches a spring 1.5 inches in equilibrium. The weight is initially displaced 8 inches above equilibrium and given a downward velocity of 4 ft/s. Find its displacement for t > 0

Answer 1

To find the displacement of a weight on a spring over time, one should use the formulas related to Simple Harmonic Motion and Hooke's Law, and apply the given initial conditions to the general SHM equation.

Step-by-step explanation:

The student's question involves finding the displacement of a weight attached to a spring over time, following the principles of Simple Harmonic Motion (SHM). Given that a weight stretches the spring by 1.5 inches in equilibrium, a displacement of 8 inches from this point, and an initial velocity of 4 ft/s downwards, the student is seeking the displacement equation for t > 0. This scenario assumes that the weight encounters no resistive forces, such as friction or air resistance, and oscillates about the equilibrium position according to Hooke's Law and SHM.

To solve this problem in physics, formulas for SHM involving Hooke's Law which states F = -kx, and Newton's second law of motion will be used. Here, k is the spring constant, and x is the displacement from the equilibrium position. The motion can also be described by equations involving amplitude, angular frequency, and phase constants in the form y(t) = Acos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase constant. To find the specific solution for this question, initial conditions, such as initial displacement and velocity, should be applied to the general solution.

Answer 2

x = 0.667 cos 10.69 t + 0.383)

Step-by-step explanation:

This is a simple harmonic movement exercise, the general solution is

x = A cos (wt + Ф)

w² = k / m

In this exercise, the initial displacement is that it corresponds to the amplitude of the movement.

A = 8 in. (1 ft. 12 in.) = ⅔ ft. = 0.6667 ft.

x = 1.5 in = 0.125 ft

We look for the constant k with Hooke's law

F = -kx

the force applied is the weight of the body F = W = mg

-mg = - k x

k = m g / x

we substitute

k = m 32.16 / 0.125

k = 257.28 m

we substitute

w = √257.28

w = 16.04 rad / s

we substitute in the expression of the position

x = 0.667 cos (16.04 t + Ф)

to find the constant fi we look for the velocity

v = dx / dt

v = - 0.667 16.04 sin (16.04 t +Ф)

indicate the values ​​of v = 4 ft / s for t = 0

- 4 = -10.6934 sin Ф

remember angles are in radians

Ф = sin⁻¹ (4/106934)

Ф = 0.383

we substitute

x = 0.667 cos 10.69 t + 0.383)