Answer:
0.059 m
Explanation:
[tex]0.5k A^2 = 0.5 (k x^{2} + m v^{2})[/tex] where k is the spring constant, A is the amplitude, x is the extension of spring, m is the mass of the spring and v is the velocity in m/s. Making a the subject then
[tex]A = \sqrt{(x^{2} + (\frac {mv^{2}}{k})}[/tex]
By substituting the given values then where m is 0.12 Kg, x= 0.0041 m, v is 0.21 m/s and K is 3 N/m then
[tex]A = \sqrt{(0.0041^{2} + (\frac {0.12\times 0.21^{2}}{3})}=0.058694122m\approx 0.059 m[/tex]
