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  • A Particle of mass  executes simple harmonic oscillations along the x-axis, with

A Particle of mass 0.8 \mathrm{~kg} executes simple harmonic oscillations along the x-axis, with a frequency of

\frac{75}{\pi} \mathrm{KZ}. At the position x=0.06 \mathrm{~m}, the Particle has Kinetic energy of 0.7 \mathrm{~J} and potential energy 0.4 \mathrm{~J}. Then amplitude of oscillation

Option: 1

0.013


Option: 2

0.012


Option: 3

0.011


Option: 4

0.11


Answers (1)

best_answer

$$ \begin{aligned} & T=2 \bar{n} \sqrt{\frac{m}{k}} \\ & \therefore f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \\ & \frac{75}{\pi}=\frac{1}{2 \pi} \sqrt{\frac{K}{0.8}} \\ & (150)^{2} \times 0.8=K \\ & K=150 \times 150 \times 0.8=\frac{150 \times 159 \times 8}{10}=150 \times 120 \\ & =18000 \mathrm{~N} / \mathrm{m} \end{aligned}

Total energy =\frac{1}{2} K A^{2}

\begin{aligned} & K E+P E=\frac{1}{2} K A 2 \\ & 0.7+0.4=\frac{1}{2} \times 18000 A^{2} \\ & \frac{1.1 \times 2}{18000}=A^{2} \\ & A=\sqrt{\frac{11 \times 2}{1800 \times 100}}=\frac{\sqrt{11}}{30 \times 10}=\frac{3.32}{300}=0.011 \end{aligned}

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Irshad Anwar

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