Question

problem 5: a child is playing on a playground spring based ride. the ride is designed to carry a 25 kg and swing the kid back and forth with a period of 0.75 s. find the stiffness of the spring used in the design (k=?).

Explanation:

Step1: Recall the formula for the period of a spring - mass system

The formula for the period \(T\) of a spring - mass system is \(T = 2\pi\sqrt{\frac{m}{k}}\), where \(m\) is the mass and \(k\) is the spring constant (stiffness). We need to solve for \(k\).
First, square both sides of the formula: \(T^{2}=(2\pi\sqrt{\frac{m}{k}})^{2}\).
Using the property \((ab)^{2}=a^{2}b^{2}\), we get \(T^{2}=4\pi^{2}\frac{m}{k}\).
Then, cross - multiply to isolate \(k\): \(k = 4\pi^{2}\frac{m}{T^{2}}\).

Step2: Substitute the given values

Given \(m = 25\space kg\) and \(T=0.75\space s\).
Substitute these values into the formula for \(k\):
\(k = 4\pi^{2}\frac{25}{(0.75)^{2}}\).
We know that \(\pi\approx3.14\), so \(4\pi^{2}\approx4\times(3.14)^{2}=4\times9.8596 = 39.4384\).
\((0.75)^{2}=0.5625\).
Then \(k=\frac{39.4384\times25}{0.5625}=\frac{985.96}{0.5625}\approx1753\space N/m\).

Answer:

The stiffness of the spring \(k\approx1753\space N/m\)