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A Particle of mass $0.6 \mathrm{~kg}$ executes simple harmonic oscillations along the x-axis, with a frequency of along the Position $x=0.08 \mathrm{~m}$ , the Particle has Kinetic energy of $0.9 \mathrm{~J}$ , and Potential energy $0.7 \mathrm{~J}$ . Then amplitude of oscillation (in m).
Option: 1
0.0345

Option: 2
0.0298

Option: 3
0.0324

Option: 4
0.0657

Answers (1)

$T=2 \pi \sqrt{\frac{m}{k}} \\$

$\therefore f=\frac{1}{2 n} \sqrt{\frac{K}{m}} \\$

$\frac{35}{\pi}=\frac{1}{2 \pi} \sqrt{\frac{K}{0.6}} \\$

$(70)^2(0.6)=K \\$

$K=2940 \mathrm{~N} / \mathrm{m}$

Total energy $=\frac{1}{2} K A^2 \\$

$K E+P E=\frac{1}{2} K A^2$

$0.9+0.4=\frac{1}{2} \times 2940 \times A^2 \\$

$2 \times 1.3=2940 \times A^2 \\$

$A^2=\frac{2 \times 1.3}{2940} \\$

$A=\sqrt{\frac{2 \times 1.3}{2940}} \\$

$A=\sqrt{\frac{26}{29400}}=\sqrt{\frac{13}{147 \times 100}}=\frac{\sqrt{13}}{\sqrt{147} \times 10} \\$ $A=\frac{3.605}{12.124 \times 10}=\frac{3605}{12124 \times 10}$

$A=0.0298$

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Kshitij

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