A 150 gram mass is attached to a horizontally aligned spring on a frictionless surface. A force of 10 Newtons will stretch the spring 25 centimeters. If the spring is compressed to 15 centimeters and then released, calculate: the spring constant; the frequency and period of the system; and the position and speed of the mass one minute after it is released.
F = kx
10 newtons = k * 0.25 meters
so
k = 40 Newtons/meter
Now do the problem
x = -0.15 at t = 0 and v = 0
so assume x = -0.15 cos (2 pi t/T)
then v = 0.15(2 pi/T) sin (2 pi t/T)
and
a = 0.15 (2pi/T)^2 cos (2pi t/T) = -(2 pi/T)^2 x
max F = m * max a
k (0.15) = m (2 pi/T)^2 (0.15
(2 pi/T)^2 = k/m = (40 Newtons/meter )/0.150 kg
f = 1/T
so
(2 pi f )^2 = 40/0.150
Now put one minute (t = 60 seconds) into the x equation
x = -0.15 cos (2 pi t/T)
To calculate the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. Mathematically, this can be written as:
F = -kx
where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, we know that a force of 10 Newtons stretches the spring by 25 centimeters. We can convert this information to SI units by using:
1 Newton = 1 kg * m/s^2
1 centimeter = 0.01 meters
So, the displacement will be:
x = 25 centimeters = 0.25 meters
Now we can rearrange Hooke's Law to solve for the spring constant:
k = -F / x
k = -10 N / 0.25 m
k = -40 N/m (Note: Negative sign indicates that the force and displacement are in opposite directions)
Therefore, the spring constant is 40 N/m.
Next, we can calculate the frequency of the system. The frequency (f) of a mass-spring system can be found using the formula:
f = 1 / (2π √(m / k))
where m is the mass and k is the spring constant.
Given the mass of 150 grams, we need to convert it to kilograms:
m = 150 grams = 0.15 kg
Now we can substitute the values into the formula:
f = 1 / (2π √(0.15 kg / 40 N/m))
Solving for f, we get:
f ≈ 0.707 Hz
The period (T) of the system is the reciprocal of the frequency:
T = 1 / f
T ≈ 1.414 seconds
Now let's calculate the position and speed of the mass one minute after it is released.
First, convert one minute to seconds:
60 seconds = 1 minute
The position of the mass at any time t in simple harmonic motion can be determined using the equation:
x(t) = A * cos(ωt + φ)
where A is the amplitude, ω is the angular frequency (ω = 2πf), t is the time, and φ is the phase constant.
We know that the amplitude (A) is the maximum displacement of the mass from its equilibrium position. In this case, the spring is compressed to 15 centimeters before being released, so the amplitude is:
A = 15 centimeters = 0.15 meters
The angular frequency (ω) can be calculated using:
ω = 2πf = 2π * 0.707 Hz
Now, substitute the values into the equation:
x(t) = 0.15 * cos(2π * 0.707 Hz * t + φ)
Since the mass is released from rest, the initial phase φ is 0.
Next, we want to find the value of x(t) after one minute. Substitute t = 60 seconds into the equation:
x(60) = 0.15 * cos(2π * 0.707 Hz * 60 seconds)
Calculating this value will give you the position of the mass after one minute.
To find the speed at that position, we can take the derivative of x(t) with respect to time. The derivative of cosine is negative sine, so the equation for the speed (v(t)) is:
v(t) = -A * ω * sin(ωt + φ)
Substituting the values we have for A, ω, and φ, we can calculate v(t) at t = 60 seconds.
This will give you the position and speed of the mass one minute after it is released.