As the kinetic energy at any displacement xxx is 12k(A2−x2)\frac{1}{2}k(A^2-x^2)21k(A2−x2)
So, 12mv2=12k(A2−x2)\frac{1}{2}mv^2=\frac{1}{2}k(A^2-x^2)21mv2=21k(A2−x2)
Hence, v=0.95697msecv=0.95697\frac{m}{sec}v=0.95697secm
Kinetic energy at x=2.80cmx=2.80cmx=2.80cm
Kinetic energy=12k(A2−x2)=0.01219J=\frac{1}{2}k(A^2-x^2)=0.01219J=21k(A2−x2)=0.01219J
Potential energy at x=2.80cmx=2.80cmx=2.80cm
Potential energy=12kx2=0.01297J=\frac{1}{2}kx^2=0.01297J=21kx2=0.01297J
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