Question

A bungee jumped of mass m jumps off a brodge. Assume that the bungee cord behaves like an ideal spring of srping constant k when the jumper is at the lowest point in the motion that follows, the extention of the bungee cord is x. What is the magnitude of acceleration of the jumper in that instant?

Answer 1

Answer:

The magnitude of the acceleration is at its maximum value.

Acceleration = $\frac{-Ak}{m} (m/s)^{2}$

Explanation:

During simple harmonic motions such as the oscillation of a spring, at the two extremes of the motion, the acceleration of the mass becomes maximum. This is because this is the point where the force on the body  is maximum.

The angular velocity $\omega=\sqrt{\frac{k}{m} }$

At points of maximum extension x, the maximum acceleration is given by the formula.

$a=-A\omega^{2}$

substituting in the value for $\omega$, this implies that

$a= -A\times (\sqrt{\frac{k}{m} } )^{2}$

⇒$a=\frac{-Ak}{m} (m/s)^{2}$