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# A spring of force constant $k$ is cut into lengths of ratio $1:2:3$. They are connected in series and the new force constant is $k\text{'}$. They are connected in parallel and the force constant is $k"$. Then $k\text{'}:k"$is:

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Solution

## Step1: Given dataThe spring constant of the spring is $k$.The length of the spring is divided into three pieces of ratio $1:2:3$.The spring constant of the system of spring when they are in series is $k\text{'}$.The spring constant of the system of spring when they are in parallel is $k"$ .Step2: Spring constantThe spring constant of a spring is the force exerted on the spring of unit length.Spring constant is defined by the form, $K=\frac{F}{x}$, where, F is the force on the spring due to extension, and x is the amount of extension.So, $springcons\mathrm{tan}t\propto \frac{1}{length}$.Step3: DiagramStep4: Finding the spring constantsLet L be the length of the spring and it is cut into three segments of ratio $1:2:3$. So, the three segments are $\frac{L}{6}$, $\frac{L}{3}$, $\frac{L}{2}$.As we know $springcons\mathrm{tan}t\propto \frac{1}{length}$.So, the spring constant of the spring segments are${K}_{1}=6K$, ${K}_{2}=3K$, ${K}_{3}=2K$So, the spring constant of the spring system when they are connected in series is$K\text{'}={K}_{1}+{K}_{2}+{K}_{3}=6K+3K+2K\phantom{\rule{0ex}{0ex}}orK\text{'}=11K.......................\left(1\right)$Again, the spring constant of the spring system when connected parallel is $\frac{1}{K\text{'}\text{'}}=\frac{1}{{K}_{1}}+\frac{1}{{K}_{2}}+\frac{1}{{K}_{3}}=\frac{1}{6K}+\frac{1}{3K}+\frac{1}{2K}\phantom{\rule{0ex}{0ex}}or\frac{1}{K\text{'}\text{'}}=\frac{1+2+3}{6K}=\frac{1}{K}\phantom{\rule{0ex}{0ex}}or\frac{1}{K\text{'}\text{'}}=\frac{1}{K}.....................\left(2\right)$Step5: Finding the ratioNow, the ratio of the two given spring systems is $\frac{K\text{'}}{K\text{'}\text{'}}=\frac{11K}{K}=11\left(From1and2\right)\phantom{\rule{0ex}{0ex}}orK\text{'}:K\text{'}\text{'}=11:1$Therefore, the ratio of spring constant is $K\text{'}:K\text{'}\text{'}=11:1$.

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