JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 21)
In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ____________ rad/s when k = 20 N/m.
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Answer
10
Explanation
$$\omega = \sqrt {{{{k_{eq}}} \over \mu }} $$
$$\mu$$ = reduced mass
springs are in series connection
$${k_{eq}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$
$${k_{eq}} = {{k \times 4k} \over {5k}} = {{4k} \over 5}$$
$${k_{eq}} = {{4 \times 20} \over 5}$$ N/m = 16 N/m
$$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = {{0.2 \times 0.8} \over {0.2 + 0.8}} = 0.16$$ kg
$$\omega = \sqrt {{{16} \over {0.16}}} = \sqrt {100} = 10$$
$$\mu$$ = reduced mass
springs are in series connection
$${k_{eq}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$
$${k_{eq}} = {{k \times 4k} \over {5k}} = {{4k} \over 5}$$
$${k_{eq}} = {{4 \times 20} \over 5}$$ N/m = 16 N/m
$$\mu = {{{m_1}{m_2}} \over {{m_1} + {m_2}}} = {{0.2 \times 0.8} \over {0.2 + 0.8}} = 0.16$$ kg
$$\omega = \sqrt {{{16} \over {0.16}}} = \sqrt {100} = 10$$
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