a steel ball weighing 128 lb is suspended from a spring, whereupon the spring is stretched 2 ft from its natural length. the ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position. assuming no damping force, find an expression for
a)the position of the ball at any time.
b)the position of the bal at t=pai/12 sec
c)the circular frequency, natural frequency and period for this system
This is a "spring-mass" mass problem.
m=128 lb
L=2 ft
g=32.2 ft/sec² (accel. due to gravity)
k=mg/L=128*32.2/2=2060.8
Let y=displacement (downwards) in the same direction as g, then the differential equation that governs the motion is:
acceleration = -ky, or
y"+ky=0 where y"=d²y/dt²
The solution of the auxiliary equation is
(m^2+k)=0, or m=±i√k
and the solution to the equation is
y=C1 cos(√k t) + C2 sin(√k t)
The initial conditions are:
y(0)=-0.5, y'(0)=0.
Since
y(0)=-0.5=C1 cos(0) + C2 sin(0)
C2=0, and C1=-0.5
y'(t)=-C2√k sin(√k t) + C2 √k cos(√k t)
From
y'(0)=0=-C2√k sin(0) + C2 √k cos(0)
we conclude
C2=0 since sin(0)=0, √k≠0 and cos(0)≠0.
So the equation of motion is:
y=-(1/2)cos(√ t)
Answers to (b) and (c) can be deduced from the equation of motion.
To solve this problem, we need to make some assumptions and apply the principles of simple harmonic motion.
Assumptions:
1. The system is idealized to have no damping force, meaning there is no resistance or friction acting on the ball.
2. The spring follows Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
3. Gravitational force can be neglected.
Let's begin:
a) Position of the ball at any time (t):
The equation for the position of an object undergoing simple harmonic motion is given by:
x(t) = A * cos(ωt + φ),
where:
- x(t) is the position of the ball at time t,
- A is the amplitude (maximum displacement) of the motion,
- ω is the angular frequency,
- φ is the phase constant.
In this case:
- The amplitude (A) is given by the initial displacement, which is 6 inches = 0.5 feet.
- The angular frequency (ω) can be determined using the formula:
ω = √(k/m),
where k is the spring constant and m is the mass of the ball.
- The phase constant (φ) is typically assumed to be zero, as we are given that the ball is started in motion with no initial velocity.
To find ω, we need to calculate the spring constant (k) by considering Hooke's Law:
F = -kx,
where F is the force exerted by the spring and x is the displacement from the equilibrium position. In this case, F is the weight of the ball.
Weight = mass * acceleration due to gravity,
128 lb = m * 32.2 ft/s^2,
m = 128 lb / 32.2 ft/s^2 = 3.977 lb·s^2/ft.
Now, we can calculate k:
F = kx,
128 lb = k * 2 ft,
k = 128 lb / 2 ft = 64 lb/ft.
Now, let's calculate ω:
ω = √(k/m) = √(64 lb/ft / (3.977 lb·s^2/ft)) = √(16.09 s^(-2)) = 4.012 s^(-1).
Putting all the values together, the expression for the position of the ball at any time is:
x(t) = 0.5 ft * cos(4.012 s^(-1) * t).
b) Position of the ball at t = π/12 sec:
To find the position of the ball at t = π/12 sec, we substitute t = π/12 into the expression we derived in part (a):
x(π/12) = 0.5 ft * cos(4.012 s^(-1) * (π/12)).
Simplifying this expression will give you the exact position of the ball at t = π/12 sec.
c) Circular frequency (ω'), Natural frequency (f), and Period (T):
The angular frequency (ω) we calculated earlier is the circular frequency. The natural frequency (f) is given by:
f = ω / (2π),
and the period (T) is the reciprocal of the natural frequency:
T = 1 / f.
So, in this case, the circular frequency is ω = 4.012 s^(-1). By substituting this value, you can calculate the natural frequency (f) and the period (T) of the system.