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 constant, k = 1.16 N/m

Explanation:

It is given that,

Mass of the air track, m = 0.2 kg

Time, t = 2.6 s

Let T is the time period of the spring. The expression for the time and spring constant is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

$k=\dfrac{4\pi^2m}{T^2}$

$k=\dfrac{4\pi^2\times 0.2}{(2.6)^2}$

k = 1.16 N/m

So, the spring’s force constant is 1.16 N/m. Hence, this is the required solution.