Question

When a mass of 29 g is attached to a certain spring, it makes 20 complete vibrations in 3.1 s. What is the spring constant of the spring? Answer in units of N/m.

Answer 1

Answer:

The spring constant of the spring is 47.62 N/m

Explanation:

Given that,

Mass that is attached with the spring, m = 29 g = 0.029 kg

The spring makes 20 complete vibrations in 3.1 s. We need to find the spring constant of the spring. We know that the number of oscillations per unit time is called frequency of an object. So,

$f=\dfrac{20}{3.1}$

f = 6.45 Hz

The frequency of oscillator is given by :

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

k is the spring constant

$k=4\pi^2f^2m$

$k=4\pi^2\times (6.45)^2\times 0.029$

k = 47.62 N/m

So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.