Question 304: Oscillations

Question

The bob of simple pendulum of length '$$L$$' is released from a position of small angular displacement $$	heta$$. Its linear displacement at time '$$\mathrm{t}$$' is ( $$\mathrm{g}=$$ acceleration due to gravity)

Accepted Answer

Equation for displacement for a particle performing S.H.M. is,
$$y=A \cos \omega t=L \cos \left(\frac{2 \pi}{T} 	imes tight)$$
But time period of a simple pendulum,
$$\begin{aligned}
T & =2 \pi \sqrt{\frac{L}{g}} \
	herefore \quad y & =L \cos \left(\frac{2 \pi}{2 \pi \sqrt{\frac{L}{g}}} 	imes tight)=L \cos \left(\sqrt{\frac{g}{L}} 	imes tight)
\end{aligned}$$
$$	herefore \quad$$ The linear displacement is,
$$s=y 	heta=L 	heta \cos \left[\sqrt{\frac{g}{L}} 	imes tight]$$

Answer

$$L 	heta \cos \left[\sqrt{\frac{g}{L}} \cdot tight]$$

Answer

$$L 	heta \sin \left[2 \pi \sqrt{\frac{g}{L}} \cdot tight]$$

Answer

$$L 	heta \cos \left[2 \pi \sqrt{\frac{g}{L}} \cdot tight]$$

Answer

$$L 	heta \sin \left[\sqrt{\frac{g}{L}} \cdot tight]$$

Oscillations

The bob of simple pendulum of length '$L' is released from a position of small angular displacement \theta. Its linear displacement at time '\mathrm{t}' is ( \mathrm{g}=$ acceleration due to gravity)

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