A 170 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm against the spring, then released. A student with a stopwatch finds that 14 oscillations take 11.0 s. What is the spring constant?

k = 10.83 N/m²

Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below

ω = √k/m

But ω = 2πf, where f = frequency.

f = number of oscillations /time taken

Number of oscillations = 14, time taken = 11s

f = 14/11 = 1.27Hz.

ω = 2×22/7×1.27

ω = 7.98 rad/s.

By substituting this parameters into ω = √k/m

Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.

7.98 = √k/0.17

By squaring both sides

(7.98)² = k/ 0.17

k = (7.98)² × 0.17

k = 10.83 N/m²

so easy. just google the electromagnetic waves and it’ll pop up.