Question

An object of mass m = 8.0 kg is attached to an ideal spring and allowed to hang in the earth's gravitational field. The spring stretches 2.2 cm before it reaches its equilibrium position.

If it were now allowed to oscillate by this spring, what would be its frequency?

Answer 1

Answer:

Frequency, f = 3.35 Hz

Explanation:

It is given that,

Mass of the object, m = 8 kg

Stretching in the spring, x = 2.2 cm

When the spring is hanged up in the Earth's gravitational field, its weight is balanced by the force in the spring. So,

$mg=kx$

k is the spring constant

$k=\dfrac{mg}{x}$

$k=\dfrac{8\times 9.8}{2.2\times 10^{-2}}$            

k = 3563.63 N/m

Let f is the frequency of oscillation. Its expression is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{3563.63}{8}}$        

f = 3.35 Hz

So, the frequency of oscillation by the spring is 3.35 Hz. Hence, this is the required solution.