Question 329: Centre of Mass and Rotational Motion

Question

A simple spring has length $$l$$ and force constant $$K$$. It is cut in to two springs of length $$l_1$$ and $$l_2$$ such that $$l_1=n l_2$$($$n$$ is an integer). The force constant of spring of length $$l_1$$ is

Accepted Answer

Let, two parts of the springs have spring constants $$K_1$$ and $$K_2$$, respectively.
$$\because \quad \begin{aligned}
& l_1=n I_2 \
& k_2=n K_1
\end{aligned}$$
Now, before cutting into two parts $$l_1$$ and $$I_2$$, we have springs in series.
$$\begin{aligned}
	ext { i.e., } & & \frac{1}{K} & =\frac{1}{K_1}+\frac{1}{K_2} \
\Rightarrow & & \frac{1}{K} & =\frac{1}{K_1}+\frac{1}{n K_1} \
\Rightarrow & & \frac{1}{K} & =\frac{n+1}{n K_1} \
\Rightarrow & & K & =\frac{n K_1}{(n+1)} \
\Rightarrow & & K_1 & =\frac{K(n+1)}{n}
\end{aligned}$$

Answer

$$K(1+n)$$

Answer

$$\frac{K(n+1)}{n}$$

Answer

$$K$$

Answer

$$\frac{K}{(n+1)}$$

Centre of Mass and Rotational Motion

A simple spring has length $l and force constant K. It is cut in to two springs of length l_1 and l_2 such that l_1=n l_2(n is an integer). The force constant of spring of length l_1$ is

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