Consider an ideal spring that has an unstretched length ℓ and a spring constant k. Suppose the spring is attached to a mass m that lies on a horizontal frictionless surface. The spring-mass system is compressed a distance of x_0 from equilibrium and then released with an initial speed v_0 toward the equilibrium position.
What is the maximum distance the spring will be stretched?
I tried conservation of energy but v_0 can't be used in the solution so I'm at a lost as to what to do.
x = a sin w t + b cos wt
v = a w cos wt - b w sin w t
omega = w = sqrt(k/m)
at t = 0
x = -xo
v = dx/dt = vo
- xo = b
+ vo = a w or a = vo/w
so x = (vo/w) sin w t - xo cos wt
now I could say
magnitude (or amplitude) = sqrt (vo^2/w^2 + xo^2) which is max motion from equilibrium
= sqrt (vo^2 m/k + xo^2)
but how to I eliminate v_0?
never mind I typed it wrong. Thanks!
To find the maximum distance the spring will be stretched, we need to consider the conservation of mechanical energy in the system.
Initially, the system is compressed a distance of x_0 from equilibrium and released with an initial speed v_0. At this point, the potential energy stored in the spring is maximum, while the kinetic energy is zero.
When the spring reaches its maximum stretch, all the initial potential energy is converted into kinetic energy. Both the potential energy and kinetic energy are given by the following formulas:
Potential Energy (U) = 0.5 * k * x^2
Kinetic Energy (K) = 0.5 * m * v^2
At maximum stretch, when the velocity is momentarily zero, the entire potential energy is converted into potential energy stored in the spring. So, we have:
Potential Energy = 0.5 * k * (maximum stretch)^2
To find the maximum stretch, we equate the initial potential energy to the potential energy at maximum stretch:
0.5 * k * x_0^2 = 0.5 * k * (maximum stretch)^2
Simplifying the equation:
x_0^2 = (maximum stretch)^2
x_0 = maximum stretch
Therefore, the maximum distance the spring will be stretched is equal to the initial compression distance, x_0.