Respuesta :
Answer:
The period of the resulting oscillatory motion is 0.20 s.
Explanation:
Given that,
Spring constant [tex]k= 3\ N/m^2[/tex]
We need to calculate the time period
The object is at rest and has no elastic potential but it does has gravitational potential.
If the object falls then the the gravitational potential change in to the elastic potential.
So,
[tex]mgh=\dfrac{1}{2}kh^2[/tex]
[tex]m=\dfrac{1}{2}\times\dfrac{kh}{g}[/tex]
Where,h = distance
k = spring constant
Put the value into the formula
[tex]m=\dfrac{1\times3\times3.42\times10^{-2}}{2\times9.8}[/tex]
[tex]m=5.235\times10^{-3}\ kg[/tex]
Using formula of time period
[tex]T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]T=\dfrac{1}{2\pi}\times\sqrt{\dfrac{5.235\times10^{-2}}{3}}[/tex]
[tex]T=0.20\ sec[/tex]
Hence, The period of the resulting oscillatory motion is 0.20 s.
