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(B) $1$. When a spring of force constant $k$ and length $L$ is cut into two parts of lengths $l_1$ and $l_2$ such that $l_1:l_2 = 1:2$,the force const

Vedclass · Accepted Answer

$T = 2 \pi \sqrt{\left(\frac{2 m}{9 k }\right)}$

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(B) $1$. When a spring of force constant $k$ and length $L$ is cut into two parts of lengths $l_1$ and $l_2$ such that $l_1:l_2 = 1:2$,the force constants of the parts are inversely proportional to their lengths $(k \propto 1/l)$.$2$. Let $l_1 = L/3$ and $l_2 = 2L/3$. Then $k_1 = 3k$ and $k_2 = 1.5k = 3k/2$.$3$. The springs are connected in parallel to the block. The equivalent spring constant $k_{eq}$ for parallel combination is $k_{eq} = k_1 + k_2 = 3k + 1.5k = 4.5k = 9k/2$.$4$. The period of oscillation $T$ is given by $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.$5$. Substituting $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m}{9k/2}} = 2\pi \sqrt{\frac{2m}{9k}}$.

$A$ spring of force constant $k$ is cut into two parts whose lengths are in the ratio $1:2$. The two parts are now connected as shown and a block of mass $m$ is connected to the combined spring. Find the period of oscillation performed by the block.

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