The kinetic energy of a particle executing si | vedclass

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(A) The kinetic energy $(K)$ of a particle in simple harmonic motion is given by $K = \frac{1}{2} k(A^2 - x^2)$,where $k$ is the force constant,$A$ is

Vedclass · Accepted Answer

$0.25$

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(A) The kinetic energy $(K)$ of a particle in simple harmonic motion is given by $K = \frac{1}{2} k(A^2 - x^2)$,where $k$ is the force constant,$A$ is the amplitude,and $x$ is the displacement.Given: $x = 3 \ cm = 0.03 \ m$,$A = 5 \ cm = 0.05 \ m$,and $K = 4 \ mJ = 4 	imes 10^{-3} \ J$.Substituting the values: $4 	imes 10^{-3} = \frac{1}{2} k((0.05)^2 - (0.03)^2)$.$4 	imes 10^{-3} = \frac{1}{2} k(0.0025 - 0.0009) = \frac{1}{2} k(0.0016) = 0.0008 k$.$k = \frac{4 	imes 10^{-3}}{8 	imes 10^{-4}} = 5 \ N/m$.The maximum force acting on the particle is $F_{max} = kA$.$F_{max} = 5 	imes 0.05 = 0.25 \ N$.

The kinetic energy of a particle executing simple harmonic motion at a displacement of $3 \ cm$ from the mean position is $4 \ mJ$. If the amplitude of the particle is $5 \ cm$,then the maximum force acting on the particle is (in $N$)

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