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23. Force position relation of particle executing SHM on \( x \)-axis is \( F+2 x-1=0 \) dyne, where \( x \) in centimeter. Mass of particle is \( 8 gm \), then find out :(i) Mean position(ii) Time period of one oscillation |
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Answer» F = - 2x + 1 F = -2 \((x - \frac{1}{2})\) F = \(-2 \times 10^{-5} \times 10^{-2}(x - \frac{1}{2})\) F = \(-2 \times 10^{-7}(x - \frac{1}{2})\) F = - kx k = \(2 \times 10^{-7}\) We know that \(\omega = \sqrt{\frac{k}{m}}\) \(\omega = \sqrt{\frac{2 \times 10^{-7} \times 1000}{0.008}}\) \(\omega = \sqrt{0.25 \times 10^{-4}}\) \(\omega = 0.5 \times 10^{-2}\) rad/sec (i) Mean position zero. (ii) Time period \(T = \frac{2\pi}{\omega}\) \(=\frac{2\pi}{0.5 \times 10^{-2}}\) \(T = \frac{20\pi \times 10^2}{5}\) \(T = 4\pi \times 10^2 \, sec\) |
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