Respuesta :
The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to [tex]\sqrt{\dfrac{m}{k} }[/tex].
Response:
- The ratio of the period to [tex]\sqrt{\dfrac{m}{k} }[/tex] is always approximately 2·π : 1
How is the value of the ratio of the period to [tex]\sqrt{\dfrac{m}{k} }[/tex] calculated?
Given:
The relationship between the period, T, the spring constant k, and the
mass attached to the spring m is presented as follows;
[tex]T = \mathbf{2 \cdot \pi \cdot \sqrt{\dfrac{m}{k} }}[/tex]
Therefore, the fraction of of the period to [tex]\sqrt{\dfrac{m}{k} }[/tex], is given as follows;
[tex]\mathbf{\dfrac{T}{ \sqrt{\dfrac{m}{k} }}} = 2 \cdot \pi[/tex]
2·π ≈ 6.23
Therefore;
[tex]T :{ \sqrt{\dfrac{m}{k} }} = 2 \cdot \pi : 1[/tex]
Which gives;
- The ratio of the period to [tex]\sqrt{\dfrac{m}{k} }[/tex] is always approximately 2·π : 1
Learn more about the oscillations in spring here:
https://brainly.com/question/14510622
