A particle of mass 0.4 text{ kg} executes sim | vedclass

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(D) Given: Mass of the particle $m = 0.4 	ext{ kg}$,Amplitude $A = 0.4 	ext{ m}$,Initial phase $\phi = \pi / 4$.The kinetic energy at the mean posit

Vedclass · Accepted Answer

$x=0.4 \sin \left((2 \sqrt{2}) t+\frac{\pi}{4}\right)$

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(D) Given: Mass of the particle $m = 0.4 	ext{ kg}$,Amplitude $A = 0.4 	ext{ m}$,Initial phase $\phi = \pi / 4$.The kinetic energy at the mean position is equal to the total energy of the oscillator,given by $KE = \frac{1}{2} m \omega^2 A^2$.Substituting the given values: $256 	imes 10^{-3} = \frac{1}{2} 	imes 0.4 	imes \omega^2 	imes (0.4)^2$.$0.256 = 0.2 	imes \omega^2 	imes 0.16$.$0.256 = 0.032 	imes \omega^2$.$\omega^2 = \frac{0.256}{0.032} = 8$.$\omega = \sqrt{8} = 2 \sqrt{2} 	ext{ rad/s}$.The general equation for simple harmonic motion is $x = A \sin(\omega t + \phi)$.Substituting the values,we get $x = 0.4 \sin \left((2 \sqrt{2}) t + \frac{\pi}{4}ight)$.

$A$ particle of mass $0.4 \text{ kg}$ executes simple harmonic motion of amplitude $0.4 \text{ m}$. When it passes through the mean position,its kinetic energy is $256 \times 10^{-3} \text{ J}$. If the initial phase of the oscillation is $\pi / 4$,then the equation of its motion is

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