A particle of mass $0.4 \mathrm{~kg}$ executes simple harmonic motion of amplitude $0.4 \mathrm{~m}$. When…

A particle of mass $0.4 \mathrm{~kg}$ executes simple harmonic motion of amplitude $0.4 \mathrm{~m}$. When it passes through the mean position, its kinetic energy is $256 \times 10^{-3} \mathrm{~J}$. If the initial phase of the oscillation is $\pi / 4$, then the equation of its motion is
  1. $x=0.4 \sin \left((0.4) t+\frac{\pi}{4}\right)$
  2. $x=02 \sin \left(2 \sqrt{2}+\left(\frac{\pi}{4}\right) t\right)$
  3. $x=0.8 \sin \left((2 \sqrt{2}) t+\frac{\pi}{2}\right)$
  4. $x=0.4 \sin \left((2 \sqrt{2}) t+\frac{\pi}{4}\right)$

Solution

Given, mass of the particle, $m=0.4 \mathrm{~kg}$ Amplitude, $A=0.4 \mathrm{~m}$ Initial phase, $\phi=\frac{\pi}{4}$ Kinetic energy at mean position $\mathrm{KE}$ As, $ \begin{aligned} & =256 \times 10^{-3} \mathrm{~J} \\ \mathrm{KE} & =\frac{1}{2} m \omega^2 A^2 \end{aligned} $ where, $\omega$ is angular frequency. $ \begin{array}{cc} \Rightarrow & \frac{1}{2} m \omega^2 A^2=256 \times 10^{-3} \\ \Rightarrow & \frac{1}{2} \times 0.4 \times \omega^2 \times(0.4)^2=256 \times 10^{-3} \\ \Rightarrow & \omega^2=8 \\ \Rightarrow & \omega=2 \sqrt{2} \mathrm{rad} \mathrm{s}^{-1} \end{array} $ Equation of simple harmonic motion is given by $ \begin{aligned} x & =A \sin (\omega t+\phi) \\ & =0.4 \sin \left[(2 \sqrt{2}) t+\frac{\pi}{4}\right] \end{aligned} $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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