Question

A 0.335 kg mass is attached to a spring and executes simple harmonic motion with a period of 0.38 s. The total energy of the system is 2.1 J. Find the force constant of the spring. Answer in units of N/m.

Answer 1

Answer:

$k=91.54 \frac{N}{m}$

Explanation:

The angular speed is defined as the angle traveled in one revolution over the time taken to travel it, that is, the period. Therefore, it is given by:

$\omega=\frac{2\pi}{T}\\\omega=\frac{2\pi}{0.38s}\\\omega=16.53\frac{rad}{s}$

The angular frequency of the simple harmonic motion of the mass-spring system is defined as follows:

$\omega=\sqrt{\frac{k}{m}}$

Here, k is the spring's constant and m is the mass of the body attached to the spring. Solving for k:

$\omega^2=\frac{k}{m}\\k=m\omega^2\\k=0.335kg(16.53\frac{rad}{s})^2\\k=91.54 \frac{N}{m}$