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(C) For a spring-block system undergoing simple harmonic motion, the total mechanical energy is conserved.The total energy $E$ is given by $E = \frac{

Vedclass · Accepted Answer

$0.275$

Vedclass · Answer

(C) For a spring-block system undergoing simple harmonic motion, the total mechanical energy is conserved.The total energy $E$ is given by $E = \frac{1}{2} k A^2$, where $k = 100 	ext{ Nm}^{-1}$ and amplitude $A = 8 	ext{ cm} = 0.08 	ext{ m}$.The potential energy at displacement $x$ is $U = \frac{1}{2} k x^2$, where $x = 3 	ext{ cm} = 0.03 	ext{ m}$.The kinetic energy $K$ at displacement $x$ is given by $K = E - U = \frac{1}{2} k (A^2 - x^2)$.Substituting the values:$K = \frac{1}{2} 	imes 100 	imes ((0.08)^2 - (0.03)^2)$$K = 50 	imes (0.0064 - 0.0009)$$K = 50 	imes 0.0055$$K = 0.275 	ext{ J}$

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