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,
[tex]f=\dfrac{20}{3.1}[/tex]
f = 6.45 Hz
The frequency of oscillator is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
k is the spring constant
[tex]k=4\pi^2f^2m[/tex]
[tex]k=4\pi^2\times (6.45)^2\times 0.029[/tex]
k = 47.62 N/m
So, the spring constant of the spring is 47.62 N/m. Hence, this is the required solution.
