Question

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

Answer 1

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 \(T = \frac{2\pi}{\omega}\)

\(=\frac{2\pi}{0.5 \times 10^{-2}}\)

\(T = \frac{20\pi \times 10^2}{5}\)

\(T = 4\pi \times 10^2 \, sec\)