Question

In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s.Find the spring's force constant.

Answer 1

Answer:

Spring force constant = 1.168 N/m

Explanation:

The period of an oscillating spring can be obtained using the formula:

T = 2π$\sqrt{\frac{m}{k} }$

The parameters available for the solution of the problem are itemized as follows:

Mass of air-track glider = 0.2 kg

Period of oscillation , T = 2.60 seconds

( This is the period of oscillation because it is the time it takes the glider to move through the equilibrium point twice, thus the time it takes for the motion to e a complete cycle)

Spring constant, k = ?

inserting the parameters, we have

2.6 = 2π$\sqrt{\frac{0.2}{k} }$

$\frac{2.6}{2pi}$ = $\sqrt{\frac{0.2}{k} }$

Squaring both sides to remove the square root, we have

$\frac{6.76}{39.48}$ = $\frac{0.2}{k}$

Solving for K, by cross multiplying, we obtain the value of k as 1.168 N/m

∴ The value of the spring's force constant is = 1.168 N/m