Answer:
Question:
What is the maximum distance the spring will be stretched
The maximum distance A the spring will be stretched is [tex]\sqrt{x_0^2+\frac{m}{k}v_0^2 }[/tex].
Explanation:
We note that for an ideal spring
F = ma = -k·x
That is m×[tex]\ddot {x}[/tex] = - k×x
If the equation above has the following initial conditions
x(0) = -x₀
[tex]\dot x (0)[/tex] = v₀
we have [tex]x(t) =-A\cdot \cos (\sqrt{\frac{k}{m} } t+\phi )[/tex]
Where:
A = Amplitude
∅ = Phase Angle
A and ∅ are such that
[tex]x_0 = A\cdot \cos (\phi )[/tex]
[tex]x(t) =A\cdot \sqrt{\frac{k}{m} } \sin \phi = v_0[/tex]
Therefore tan ∅ = [tex]\frac{v_0}{x_0} \sqrt{\frac{m}{k} }[/tex] and
Also, the amplitude A = Maximum displacement from equilibrium or the maximum distance the spring will be stretched is given by;
A = [tex]\sqrt{x_0^2+\frac{m}{k}v_0^2 }[/tex].
