Question

A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat its frequency of vibration is f= 3.0 Hz

Answer 1

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

$mg=kx$

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

$k=\dfrac{2.8\times 9.8}{0.018}$

k = 1524.44 N/m

Since, $\omega=\sqrt{\dfrac{k}{m}}$

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

$m=\dfrac{1524.44}{4\pi^2 \times 3^2}$

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.