Question 341: Centre of Mass and Rotational Motion

Question

Two massless springs of spring constant $$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ are connected one after the other forming a single chain, suspended vertically and certain mass is attached to the free end. If '$$e_1$$' and '$$e_2$$' are their respective extensions and '$$\mathrm{f}$$' is their stretching force, the total extension produced is

Accepted Answer

For a spring $$\mathrm{F}=\mathrm{Kx}$$
$$\begin{aligned}
& 	herefore \mathrm{x}=\frac{\mathrm{F}}{\mathrm{K}} \
& 	herefore \mathrm{x}_1=\frac{\mathrm{F}}{\mathrm{K}_1} 	ext { and } \mathrm{x}_2=\frac{\mathrm{F}}{\mathrm{K}_2} \
& 	ext { or } \mathrm{e}_1=\frac{\mathrm{F}}{\mathrm{K}_1} 	ext { and } \mathrm{e}_2=\frac{\mathrm{F}}{\mathrm{K}_2}
\end{aligned}$$
$$	herefore \mathrm{e}_1+\mathrm{e}_2=\mathrm{F}\left(\frac{1}{\mathrm{~K}_1}+\frac{1}{\mathrm{~K}_2}ight)$$
Note: The springs are connected in series.
$$	herefore$$ The effective spring constant $$\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_1+\mathrm{K}_2}$$
$$	herefore$$ Total extension
$$\mathrm{e}=\frac{\mathrm{F}}{\mathrm{K}}=\frac{\mathrm{F}\left(\mathrm{K}_1+\mathrm{K}_2ight)}{\mathrm{K}_1 \mathrm{~K}_2}=\frac{\mathrm{F}}{\mathrm{K}_2}+\frac{\mathrm{F}}{\mathrm{K}_1}$$

Answer

$$\mathrm{f}\left(\frac{1}{\mathrm{~K}_1}+\frac{1}{\mathrm{~K}_2}\right)$$

Answer

$$\mathrm{f}\left(\frac{1}{\mathrm{~K}_1}-\frac{1}{\mathrm{~K}_2}\right)$$

Answer

$$\mathrm{f}\left(\mathrm{K}_1+\mathrm{K}_2\right)$$

Answer

$$\mathrm{f}\left(\mathrm{K}_1-\mathrm{K}_2\right)$$

Centre of Mass and Rotational Motion

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