JEE MAIN - Physics (2021 - 17th March Morning Shift - No. 25)
Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is Tb/Ta = $$\sqrt x $$, where value of x is ___________. (Round off to the Nearest Integer)
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Answer
2
Explanation
$${T_a} = 2\pi \sqrt {{M \over k}} $$
$${K_{e{q_{series}}}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}} = {k \over 2}$$
$${T_b} = 2\pi \sqrt {{M \over {k/2}}} = 2\pi \sqrt {{{2M} \over k}} $$
$${T_b} = \sqrt 2 {T_a}$$
$${{{T_b}} \over {{T_a}}} = \sqrt 2 $$
$$ \therefore $$ x = 2
$${K_{e{q_{series}}}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}} = {k \over 2}$$
$${T_b} = 2\pi \sqrt {{M \over {k/2}}} = 2\pi \sqrt {{{2M} \over k}} $$
$${T_b} = \sqrt 2 {T_a}$$
$${{{T_b}} \over {{T_a}}} = \sqrt 2 $$
$$ \therefore $$ x = 2
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