Physics - SHM Question with Solution | TestHub

Equation for displacement for a particle performing S.H.M. is, $$\mathrm{y}=\mathrm{A}\cos \omega \mathrm{t}=\mathrm{L}\cos \left(\frac{2\pi }{\mathrm{T}}\times \mathrm{t}\right)$$ But time period of a simple pendulum, $$\mathrm{T}=2\pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$$ $\therefore y=L\cos \left(\frac{2\pi }{2\pi \sqrt{\frac{L}{g}}\times t}\right)=L\cos \left(\sqrt{\frac{g}{L}}\times t\right)$ $\therefore$ The linear displacement is, $$\mathrm{s}=\mathrm{y}\theta =L\theta \cos \left[\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\times \mathrm{t}\right]$$