A mass m is at rest on the end of a spring of spring constant k. At t = 0 it is given an impulse J by a hammer. Write the formula for the subsequent motion in terms of m, k, J, and t. Assume x=0 at t=0.
Thanks,
-nik
To derive the formula for the subsequent motion of the mass, we can use Newton's second law of motion and the equation for the force exerted by a spring.
1. Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In this case, the net force is provided by the spring force, which is given by Hooke's Law.
2. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula for the spring force is F = -kx, where k is the spring constant and x is the displacement.
3. By applying Newton's second law to this situation, we have the equation m*a = -k*x, where m is the mass and a is the acceleration.
4. The acceleration can be obtained by taking the second derivative of the displacement with respect to time, a = d^2x/dt^2. Similarly, the displacement x is the first integral of the velocity, x = ∫v*dt, and the velocity v is the first derivative of the displacement, v = dx/dt.
5. Substituting these expressions into the equation m*a = -k*x, we have m*(d^2x/dt^2) = -k*x.
6. Rearranging the equation, we get d^2x/dt^2 = -(k/m) * x.
This is a second-order linear homogeneous differential equation with constant coefficients. The solution to this equation takes the form of x(t) = A*cos(ω*t) + B*sin(ω*t), where A and B are arbitrary constants, and ω is the angular frequency given by ω = √(k/m).
7. At t = 0, the initial conditions state that the mass is at rest, so the velocity is zero, v(0) = 0. Differentiating the equation for x(t), we find v(t) = -(A*ω)*sin(ω*t) + (B*ω)*cos(ω*t). Substituting t = 0 yields v(0) = B*ω = 0, leading to B = 0.
8. Therefore, the formula for the subsequent motion of the mass in terms of m, k, J, and t is x(t) = A*cos(ω*t), where A is the amplitude determined by the initial impulse J, and ω = √(k/m) is the angular frequency.
To determine the subsequent motion of the mass m after being given an impulse J by a hammer, we need to consider the equation of motion for the system.
The equation of motion for a mass-spring system is given by:
m * d²x/dt² + k * x = 0
where m is the mass of the object, x is the displacement of the object from its equilibrium position, t is the time, and k is the spring constant.
In this case, since the mass is initially at rest, we can assume that the initial displacement x = 0 and the initial velocity v = 0. Therefore, the equation of motion can be simplified to:
m * d²x/dt² + k * x = J * δ(t)
where δ(t) is the Dirac delta function, representing the impulse J applied at t=0.
To solve this differential equation, we will need to determine the general solution for the homogeneous equation (without the impulse term) and then add the particular solution representing the impulse.
The general solution for the homogeneous equation is of the form:
x(t) = A * cos(ωt) + B * sin(ωt)
where A and B are constants to be determined and ω = sqrt(k/m) is the angular velocity of the system.
Now, to find the particular solution representing the impulse, we can use the fact that the impulse causes an instantaneous change in velocity. Since the mass is initially at rest, we have:
v(0+) - v(0-) = J/m
where v(0+) is the velocity immediately after the impulse and v(0-) is the velocity just before the impulse. Since the mass is initially at rest, we can assume v(0-) = 0, so:
v(0+) = J/m
Differentiating x(t) with respect to time, we obtain:
v(t) = -A * ω * sin(ωt) + B * ω * cos(ωt)
Substituting t = 0 into this expression, we get:
v(0+) = -A * ω * sin(0) + B * ω * cos(0) = B * ω = J/m
Therefore, B = J/(m * ω).
Substituting these values of A and B back into the general solution, we get:
x(t) = A * cos(ωt) + (J/(m * ω)) * sin(ωt)
Now, to determine the constant A, we can use the initial condition x(0) = 0. Substituting t = 0 into the equation, we get:
x(0) = A * cos(0) + (J/(m * ω)) * sin(0) = A = 0
Therefore, the final formula for the subsequent motion of the mass m in terms of m, k, J, and t is:
x(t) = (J/(m * ω)) * sin(ωt)
where ω = sqrt(k/m).
This formula describes how the position of the mass varies with time after being given the impulse J at t = 0.