Explanation:
Given that,
Mass of block = 2.00 kg
Spring constant = 10.0 N/m
Initial amplitude [tex]x_{i}= 25.0[/tex]
Final amplitude [tex]x_{f}= 0.75 x_{i}[/tex]
Number of oscillation = 4
(a). We need to calculate the time
Given formula,
[tex]t = nT[/tex]
[tex]t=n2\pi\sqrt{\dfrac{m}{k}}[/tex]
Put the value into the formula
[tex]t=4\times2\pi\sqrt{\dfrac{2.00}{10.0}}[/tex]
[tex]t=11.23\ sec[/tex]
We need to calculate the value of b
Using damping formula
[tex]x_{m}e^{\dfrac{-bt}{2m}}=0.75x_{i}[/tex]
[tex]e^{\dfrac{-b\times11.23}{2\times2.00}}=0.75[/tex]
[tex]e^{-2.81b}=0.75[/tex]
[tex]-2.81 b=ln(0.75)[/tex]
[tex]b=\dfrac{0.28}{2.81}[/tex]
[tex]b=0.099\ kg/s[/tex]
(b), The damping force equation is
[tex]m\dfrac{d^2x}{dt^2}+b\dfrac{dx}{dt}+kx=0[/tex]
We know that,
The time period is
[tex]T=\dfrac{2\pi}{\sqrt{\dfrac{k}{m}+(\dfrac{b}{2m})^2}}[/tex]
Therefore, time period is affected by damping force.
Hence, This is the required solution.