Now suppose that the weight attached to the spring is pulled down a further 5cm from its equilibrium positions and released from rest. Show that newtons second law leads to the dynamic equation (in SI units)
d^2z/dt^2 + 36z - g = 0
where z(t) measures the extension of the spring from its natural lenght and g is the acceleration due to gravity.
The first part of this question was related to Hooke's Law which i have answered correctly, i am just unsure how you start this particular question.
To start this problem, we can apply Newton's second law to the weight attached to the spring. According to Newton's second law, the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is due to the weight of the object and the force exerted by the spring.
Let's denote the extension of the spring from its natural length as z(t). The force exerted by the spring is proportional to the displacement and is given by Hooke's law as F_s = -kz(t), where k is the spring constant.
The weight of the object is given by the force of gravity, which is equal to the mass of the object multiplied by the acceleration due to gravity, g.
Therefore, the net force acting on the weight attached to the spring is given by F_net = F_s + F_g, where F_g = mg.
Using Newton's second law, we have:
m(d^2z/dt^2) = -kz(t) + mg
Dividing both sides by m, we get:
d^2z/dt^2 = -kz(t)/m + g
Since we know that the angular frequency of the spring is given by ω = √(k/m), we can substitute this expression into the equation:
d^2z/dt^2 = -(ω^2)z(t) + g
Simplifying further using ω^2 = k/m, we have:
d^2z/dt^2 = -36z(t) + g
Therefore, Newton's second law leads to the dynamic equation:
d^2z/dt^2 + 36z - g = 0
where z(t) measures the extension of the spring from its natural length and g is the acceleration due to gravity.
To start this question, we need to use Newton's second law, which states that the force acting on an object is equal to its mass multiplied by its acceleration:
F = m * a
In this case, the object is the weight attached to the spring. Let's assume that the weight has mass m and the acceleration of the weight is a.
The force acting on the weight can be divided into two components: the force due to gravity and the force due to the spring. The force due to gravity is simply the weight of the object, which we can calculate by multiplying its mass by the acceleration due to gravity, g:
F_gravity = m * g
The force due to the spring can be determined by Hooke's Law, which states that the force exerted by a spring is proportional to its extension. We denote the extension of the spring as z(t), which measures the displacement from its equilibrium position. According to Hooke's Law, the force due to the spring is given by:
F_spring = -k * z
where k is the spring constant.
Note that we have a negative sign because the force exerted by the spring is in the opposite direction to its displacement.
Now, according to Newton's second law, the total force acting on the weight is the sum of the forces due to gravity and the spring:
F_total = F_gravity + F_spring
Substituting the expressions for F_gravity and F_spring, we get:
F_total = m * g - k * z
Since acceleration is the second derivative of displacement with respect to time, we can rewrite the equation as:
m * d^2z/dt^2 = m * g - k * z
Dividing both sides of the equation by m, we have:
d^2z/dt^2 = g - (k/m) * z
Recall that (k/m) is a constant, denoted as ω^2 (omega-squared), representing the angular frequency of the spring-mass system. Therefore, we can rewrite the equation as:
d^2z/dt^2 + ω^2 * z = g
Comparing this equation with the given dynamic equation:
d^2z/dt^2 + 36z - g = 0
We can conclude that ω^2 = 36 (which implies ω = 6) and the negative sign of the spring constant k has been absorbed in the definition of ω.
Thus, Newton's second law leads to the dynamic equation d^2z/dt^2 + 36z - g = 0, where z(t) represents the extension of the spring from its natural length and g is the acceleration due to gravity.