a spring is stretched 5*10^2m by a force of 5*10^-4N.a mass of 0.001kg is placed on the lower end of the spring.after equilibrium has been reached, the upper end of the spring is moved up and down so that the external force acting on the mass is given by F(t)=20coswt.calculate the position of the mass at any time , measured from equilibrium position and the angular frequency for which resonance occurs.
To find the position of the mass at any time, we need to consider the equation of motion for a spring-mass system. The equation is given by:
m * d^2x/dt^2 + k * x = F(t)
where:
- m is the mass of the object (0.001 kg),
- x is the displacement of the mass from its equilibrium position,
- k is the spring constant,
- F(t) is the external force acting on the mass.
In this case, we have F(t) = 20 * cos(wt), where w is the angular frequency.
We need to solve this differential equation to find the position of the mass at any time. First, let's determine the values of k and w.
Given that the spring is stretched 5 * 10^2 m by a force of 5 * 10^-4 N, we can use Hooke's Law to find the spring constant:
k = F / x = (5 * 10^-4 N) / (5 * 10^2 m) = 10^-6 N/m
Next, we need to determine the angular frequency w for resonance. Resonance occurs when the natural frequency of the system matches the frequency of the external force. In this case, the natural frequency (ω_n) is given by:
ω_n = √(k / m)
Substituting the values, we get:
ω_n = √(10^-6 N/m / 0.001 kg) = √(10^-3) = 10^(-3/2) rad/s
To find the angular frequency for resonance, we set ω = ω_n and solve for w:
w = ω_n = 10^(-3/2) rad/s
Now that we have the values of k and w, we can solve the differential equation to find the position of the mass at any time.
Let's assume the initial conditions are x(0) = 0 (mass at equilibrium) and v(0) = 0 (initial velocity = 0). We can also write the equation of motion as:
m * d^2x/dt^2 + k * x = 20 * cos(wt)
This is a second-order linear homogeneous differential equation with constant coefficients. The general solution to this differential equation is of the form:
x(t) = A * cos(ω_d * t) + B * sin(ω_d * t)
where A and B are constants determined by the initial conditions and ω_d is the angular frequency of the damped system.
Since we are given that the system reaches equilibrium, we can assume it is undamped and that there is no velocity initially (v(0) = 0). Therefore, we have:
ω_d = √(k/m) = ω_n = 10^(-3/2) rad/s
Hence, the position equation becomes:
x(t) = A * cos(ω_n * t) + B * sin(ω_n * t)
To determine the constants A and B, we need to apply the initial conditions. Since the mass is at equilibrium (x(0) = 0), we have:
x(0) = A * cos(0) + B * sin(0) = A * (1) + B * (0) = A = 0
Thus, the equation for the position of the mass at any time becomes:
x(t) = B * sin(ω_n * t)
Finally, we have the position formula:
x(t) = B * sin(10^(-3/2) * t)
where B is a constant that will depend on the displacement of the mass before equilibrium is reached.