Answer:
(A) Spring constant will be 126.58 N/m
(B) Amplitude will be equal to 0.177 m
Explanation:
We have given mass of the block m = 200 gram = 0.2 kg
Time period T = 0.250 sec
Total energy is given TE = 2 J
(A) For mass spring system time period is equal to [tex]T=2\pi \sqrt{\frac{m}{K}}[/tex]
So [tex]0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}[/tex]
[tex]0.0398=\sqrt{\frac{0.2}{K}}[/tex]
Now squaring both side
[tex]0.00158=\frac{0.2}{K}[/tex]
K = 126.58 N/m
So the spring constant of the spring will be 126.58 N/m
(B) Total energy is equal to [tex]TE=\frac{1}{2}KA^2[/tex], here K is spring constant and A is amplitude
So [tex]2=\frac{1}{2}\times 126.58\times A^2[/tex]
[tex]A^2=0.0316[/tex]
A = 0.177 m
So the amplitude of the wave will be equal to 0.177 m
