A mass 0.4 text{ kg} performs S.H.M. with a f | vedclass

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Question

(A) The total energy $E$ of a particle performing $S.H.M.$ is the sum of its kinetic energy $K$ and potential energy $U$.$E = K + U = 2 	ext{ J} + 1.

Vedclass · Accepted Answer

$0.125$

Vedclass · Answer

(A) The total energy $E$ of a particle performing $S.H.M.$ is the sum of its kinetic energy $K$ and potential energy $U$.$E = K + U = 2 	ext{ J} + 1.2 	ext{ J} = 3.2 	ext{ J}$.The formula for total energy in $S.H.M.$ is $E = \frac{1}{2} m \omega^2 A^2$, where $m$ is mass, $\omega$ is angular frequency, and $A$ is amplitude.Given frequency $f = \frac{16}{\pi} 	ext{ Hz}$, the angular frequency $\omega = 2 \pi f = 2 \pi \left( \frac{16}{\pi} ight) = 32 	ext{ rad/s}$.Substituting the values into the energy equation:$3.2 = \frac{1}{2} 	imes 0.4 	imes (32)^2 	imes A^2$$3.2 = 0.2 	imes 1024 	imes A^2$$3.2 = 204.8 	imes A^2$$A^2 = \frac{3.2}{204.8} = \frac{1}{64}$$A = \sqrt{\frac{1}{64}} = 0.125 	ext{ m}$.

$A$ mass $0.4 \text{ kg}$ performs $S.H.M.$ with a frequency $\frac{16}{\pi} \text{ Hz}$. At a certain displacement, it has kinetic energy $2 \text{ J}$ and potential energy $1.2 \text{ J}$. The amplitude of oscillation is (in $\text{ m}$)

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