Question

A mass weighing 32 pounds stretches a spring 2 feet. Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s. amplitude ft period s How many complete cycles will the mass have completed at the end of 6π seconds? cycles

Answer 1

Answer:

A) Amplitude = 1.803 ft ; Period = π/2 s

B) Cycles in 6π s = 12 cycles

Explanation:

Let M be the mass attached and let K be the spring constant.

Applying Newton's second law to this system, we have;

m(d²x/dt²) = -kx²

When x(t) is the displacement from the equilibrium position, the equation can be given as;

d²x/dt² + (k/m)x = 0 - - - - - (eq1)

Let's find the mass.

W=mg

W = 32lb and g = 32 ft/s²

So, m = W/g = 32/32 = 1 lbft/s² = 1 slug

Let's find spring stiffness;

From Hooke's law, W = ks

s = 2 ft. Thus, k = W/s = 32/2 = 16 lb/ft

Now, putting the values of m and k into eq 1,we obtain;

d²x/dt² + (16/1)x = 0

d²x/dt² + 16x = 0

Now, we know that the solution of x" + ω²x = 0 is given as;

x(t) = C1 cosωt + C2 sinωt

By inspection, ω² = 16. Thus, ω = 4

So, the general solution is now;

x(t) = C1 cos4t + C2 sin4t

The initial conditions from the question are;

x(0) = -1 ft and x'(0) = -6 ft/s

At x(0) = -1

-1 = C1 cos 0 + C2 sin 0

-1 = C1

Now, at x'(0) = 6

x' = -4C1sin4t + 4C2cos4t

So,

6 = -4C1sin 0 + 4C2cos0

4C2 = - 6

C2 = -6/4 = -3/2

Thus, the equation of motion is;

x(t) = - cos4t - (3/2)sin4t

A) Amplitude is given as;

A = √(C1² + C2²) = √(-1² + (-3/2)²)

A = √(1 + 9/4) = √(13/4) = 1.803 ft

Period is given as;

T = 2π/ω = 2π/4 = π/2 s

B) Cycles it will complete in 6π;

= 6π/T = 6π/(π/2) = 12 cycles