The amplitude of a particle executing SHM is | vedclass

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(B) Let the amplitude be $A = 3 	ext{ cm}$.Let the displacement be $x$.The kinetic energy $(KE)$ of a particle in $SHM$ is given by $KE = \frac{1}{2}

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$2$

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(B) Let the amplitude be $A = 3 	ext{ cm}$.Let the displacement be $x$.The kinetic energy $(KE)$ of a particle in $SHM$ is given by $KE = \frac{1}{2} k(A^2 - x^2)$.The potential energy $(PE)$ of a particle in $SHM$ is given by $PE = \frac{1}{2} kx^2$.According to the problem, $KE = PE + 0.25 PE = 1.25 PE = \frac{5}{4} PE$.Substituting the expressions for $KE$ and $PE$:$\frac{1}{2} k(A^2 - x^2) = \frac{5}{4} (\frac{1}{2} kx^2)$.$A^2 - x^2 = \frac{5}{4} x^2$.$A^2 = x^2 + \frac{5}{4} x^2 = \frac{9}{4} x^2$.Taking the square root of both sides:$A = \frac{3}{2} x$.Given $A = 3 	ext{ cm}$, we have $3 = \frac{3}{2} x$.Therefore, $x = 2 	ext{ cm}$.

The amplitude of a particle executing $SHM$ is $3 \text{ cm}$. The displacement at which its kinetic energy will be $25\%$ more than its potential energy is (in $\text{cm}$):

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