At time an object of mass 50.34 g on the end of a horizontal spring is moving to the right at 10 cm/s and is displaced 2 cm to the left from the equilibrium position. If the spring constant is 0.234N/m, find an expression for the position of the particle as a function of time.
To find an expression for the position of the particle as a function of time, we can utilize Newton's second law of motion and Hooke's law.
1. Start by determining the force acting on the object. Since the object is connected to a spring, the force is given by Hooke's law, which states that the force is directly proportional to the displacement from the equilbrium position. The force is given as:
F = -kx
where F is the force, k is the spring constant, and x is the displacement.
2. Calculate the acceleration of the object using Newton's second law of motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass. In this case, the net force is the force we calculated in step 1, and the mass is given as:
m = 50.34 g = 0.05034 kg
The acceleration is given as:
F = ma
-kx = ma
a = -kx/m
3. Integrate the acceleration with respect to time to find the velocity. We can rewrite the acceleration as:
a = d^2x/dt^2 = -kx/m
Rearranging the equation gives:
d^2x = -kx/m dt^2
Integrating both sides with respect to time gives:
∫d^2x = -k/m ∫x dt^2
∫d^2x = -k/m ∫x d(∫dt)
Using the chain rule, we can rewrite d^2x as (d^2x/dt^2) dt:
∫(d^2x/dt^2) dt = -k/m ∫x d(∫dt)
∫a dt = -k/m ∫x d(∫dt)
∫a dt = -k/m ∫x dx
Integrating both sides gives:
∫a dt = -k/m ∫x dx
∫a dt = -k/m ∫x dx
∫a dt = -k/m ∫x dx
=> ∫a dt = -k/m ∫x dx
=> ∫a dt = -k/m ∫x dx
=> ∫a dt = -k/m ∫x dx
Now, integrating the left-hand side gives:
∫a dt = ∫(d^2x/dt^2) dt = ∫d(d(x))/dt dt
∫a dt = ∫d(dx) = ∫dx
Therefore, the left-hand side of the equation becomes:
∫a dt = ∫dx = x
The right-hand side of the equation becomes:
-k/m ∫x dx = -k/m * (1/2) * x^2 = (-kx^2)/(2m)
So, we have:
x = (-kx^2)/(2m)
Rearranging the equation, we get:
2mx = -kx^2
Dividing both sides by -kx^2, we get:
2m/k = -x/dx
Integrating both sides with respect to x:
∫(2m/k) dx = ∫(-x/dx) dx
2mx/k + C1 = -x^2/2 + C2
Simplifying the equation, we obtain:
2mx/k = -x^2/2 + C
Rearranging the equation, we get:
x^2 + 4mx/k = C
The constant C can be determined using the initial conditions. We are given that the object is initially displaced 2 cm to the left (x = -2 cm) and moving to the right at 10 cm/s (dx/dt = 10 cm/s).
Plugging these values into the equation and solving for C:
(-2 cm)^2 + 4m(0.234 N/m)/k = C
4 cm^2 + 4(0.05034 kg)(0.234 N/m)/0.234 N/m = C
4 cm^2 + 4(0.05034 kg) = C
4 cm^2 + 0.20136 kg = C
C = 4.00136 cm^2 + kg
Thus, the expression for the position of the particle as a function of time is:
x^2 + 4mx/k = 4.00136 cm^2 + kg