Question

A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance x = 11 cm from its equilibrium position on a frictionless surface and released from rest at t = 0. What are the angular frequency, frequency and period of the resulting motion?

Answer 1

To solve this problem we will apply the kinematic equations of the simple harmonic movement, for which the angular velocity is expressed as a function of the square root of the spring constant and the mass. From the angular velocity it will be possible to find the frequency and finally the period - which is the inverse of the latter.

Mathematically the expression of angular movement for a spring satisfies equality:

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

Where,

k = Spring constant

m = Mass

Then the angular frequency is

$\omega = \sqrt{\frac{65}{0.680}}$

$\omega = 9.7769rad/s$

Now the frequency is defined as the ratio of change between the angular frequency and the constant 2π, that is

$f = \frac{\omega}{2\pi}$

$f = \frac{9.7769}{2\pi}$

$f = 1.55604 Hz$

The period is the inverse of the frequency, therefore

$T = \frac{1}{f}$

$T = \frac{1}{1.55604}$

$T= 0.6426s$