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
k = 64 lb/ft for spring
position above equilirium = x
at t = 0 x = .5 ft
so
x = .5 cos 2 pi t/T
we need to find T, the period
f = -kx = m a
but a = .5 (2 pi/T)^2(-cos2 pi/T) =-(2pi/T)^2 x
so
-k x = -(2 pi/T)^2 x
2 pi/T = sqrt(k/m)
m = 128/32 = 4
so
2 pi/T = sqrt (64/4) = 4
therefore
x = .5 cos 4 t
at t = pi/12
x = .5 cos pi/3 = .5 cos 60deg = .25
w = 2 pi f = 2 pi/T = 4 radians/sec
f = w/2pi = 2/pi
T = pi/2
a) Oh, look at Mr. Steel Ball over here, thinking he's so important! Well, to find the position of the ball at any time, we use Hooke's Law, which states that the force exerted by the spring is proportional to the displacement from its equilibrium position. So, the equation for the position of the ball is: x(t) = 2ft cos(sqrt(k/m) * t), where x(t) is the position at time t, f is the initial displacement in feet, k is the spring constant, and m is the mass of the ball.
b) Ah, the ball wants to know where it'll be at t=pai/12 sec? Such an irrational question! But hey, we can use the same position equation I mentioned earlier, substituting t=pai/12: x(pai/12) = 2ft cos(sqrt(k/m) * pai/12).
c) Let's talk frequencies! The circular frequency, denoted as ω, is given by ω = sqrt(k/m). The natural frequency, denoted as ω_n, is also the same as the circular frequency, which translates to ω_n = sqrt(k/m). And finally, the period, denoted as T, is the reciprocal of the natural frequency: T = 2π/ω_n.
Just remember, kids, this is all assuming no damping force, otherwise the whole party gets even crazier!
To find the position of the ball at any time, we can use the equation of motion for a spring-mass system:
mx'' + kx = 0
where m is the mass, x is the displacement, and k is the spring constant.
a) The mass of the ball is given as 128 lb, which we can convert to slugs (1 slug = 32.174 lb):
m = 128 lb / 32.174 lb/slug ≈ 3.98 slugs
The spring constant k can be determined using Hooke's Law:
k = F / x
where F is the weight of the ball (mg) and x is the displacement of the spring (2 ft = 24 in):
k = (3.98 slugs) * (32.174 ft/s^2) / (24 in) ≈ 5.31 slugs/ft
We can let x(t) be the position of the ball at any time t. The general solution to the equation of motion is given by:
x(t) = A * cos(ωt) + B * sin(ωt)
where A and B are constants that can be determined using the initial conditions.
The initial displacement is given as 6 in above the equilibrium position, which means x(0) = 6 in. Substituting these values into the equation, we get:
6 in = A * cos(0) + B * sin(0)
6 in = A * 1 + B * 0
6 in = A
So A = 6 in.
The initial velocity is not given, so we can assume it to be zero, which means x'(0) = 0. Differentiating x(t) with respect to t, we get:
x'(t) = -A * ω * sin(ωt) + B * ω * cos(ωt)
Substituting t = 0 and x'(0) = 0 into the equation, we get:
0 = -A * ω * sin(0) + B * ω * cos(0)
0 = B * ω
Since ω cannot be zero (otherwise x(t) would be constant), we have B = 0.
Therefore, the expression for the position of the ball at any time is:
x(t) = (6 in) * cos(ωt)
b) To find the position of the ball at t = π/12 sec, substitute t = π/12 into the equation:
x(π/12) = (6 in) * cos(ω * π/12)
c) The circular frequency ω can be determined from the relationship:
ω = √(k/m)
Substituting the values of k and m, we get:
ω = √(5.31 slugs/ft / 3.98 slugs)
ω = √1.3367 ft^(-1)
ω ≈ 1.156 ft^(-1)
The natural frequency (in Hz) can be found by dividing the circular frequency by 2π:
f = ω / (2π)
f ≈ 1.156 ft^(-1) / (2π) ≈ 0.183 Hz
The period (T) is the reciprocal of the frequency:
T = 1 / f ≈ 1 / 0.183 s ≈ 5.462 s
Therefore, the expression for the position of the ball at any time is (6 in) * cos(1.156 ft^(-1)t), the position of the ball at t = π/12 sec is (6 in) * cos(1.156 ft^(-1) * π/12), and the circular frequency is ω ≈ 1.156 ft^(-1), the natural frequency is f ≈ 0.183 Hz, and the period is T ≈ 5.462 s.
To solve this problem, we need to use Newton's second law of motion and Hooke's Law. Let's break down the problem step by step:
a) The position of the ball at any time:
First, let's find the spring constant (k) using Hooke's Law. Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position. Mathematically, it can be written as F = -kx, where F is the force applied, k is the spring constant, and x is the displacement.
Since the ball is suspended, its weight is balanced by the spring tension. Therefore, we can write the equation as:
mg = kx (where m is the mass of the ball and g is the acceleration due to gravity)
Given that the weight of the ball is 128 lb and the spring is stretched 2 feet, we have:
128 lb = k * 2 ft
Dividing both sides by 2 ft, we get:
k = 64 lb/ft
Now, let's find the equation for the position of the ball at any time. We'll use the equation for simple harmonic motion (SHM), which 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 of the motion (maximum displacement)
- ω is the angular frequency (also known as circular frequency)
- φ is the phase constant (initial phase angle)
In this case, the amplitude A is 6 inches, and we need to find ω and φ.
To find the angular frequency ω, we can use the formula:
ω = sqrt(k/m)
Given that the mass of the ball is 128 lb, we need to convert it to slugs (since lb is a unit of force, not mass). 1 slug is equal to a mass that accelerates at 1 ft/s^2 when a force of 1 lb is applied. Therefore:
m = 128 lb / g (where g is the acceleration due to gravity in ft/s^2)
By substituting g = 32.17 ft/s^2 (approximate value for Earth's gravity), we have:
m = 128 lb / 32.17 ft/s^2
Converting lb to slugs, we can rewrite m as:
m = 128 lb / (32.17 ft/s^2 * 32.2 lb/ft)
m = 4 lb * s^2 / ft
Now we can calculate ω:
ω = sqrt(k/m)
ω = sqrt(64 lb/ft / (4 lb * s^2 / ft))
ω = sqrt(16 s^2/s^2)
ω = 4 rad/s
Next, we need to find the phase constant, φ. Given that the ball is started in motion with no initial velocity by displacing it 6 inches above the equilibrium position, we can write:
x(0) = A * cos(φ)
Since x(0) = 6 inches, we have:
6 inches = A * cos(φ)
Dividing by A, we get:
cos(φ) = 6 inches / A
cos(φ) = 6 inches / 6 inches
cos(φ) = 1
Therefore, φ = 0.
Now we have all the values to express the position of the ball at any time:
x(t) = 6 cos(4t + 0)
Simplifying, we get:
x(t) = 6 cos(4t)
b) The position of the ball at t = π/12 sec:
To find the position of the ball at t = π/12 sec, we can substitute π/12 into the equation we derived in part (a):
x(t) = 6 cos(4t)
x(π/12) = 6 cos(4 * (π/12))
x(π/12) = 6 cos(π/3)
Using the value of cos(π/3) = 1/2, we get:
x(π/12) = 6 * (1/2)
x(π/12) = 3
Therefore, the position of the ball at t = π/12 sec is 3 inches above the equilibrium position.
c) The circular frequency (ω), natural frequency, and period:
We already found the circular frequency ω in part (a), which is ω = 4 rad/s.
The natural frequency (ω_n) is defined as the circular frequency divided by 2π:
ω_n = ω / (2π)
ω_n = 4 rad/s / (2π)
ω_n ≈ 0.6366 rad/s
Lastly, the period (T) is the time taken for one complete oscillation and is given by the reciprocal of the natural frequency:
T = 1 / ω_n
T = 1 / 0.6366 rad/s
T ≈ 1.571 s
Therefore, the circular frequency is 4 rad/s, the natural frequency is approximately 0.6366 rad/s, and the period is approximately 1.571 seconds.