JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 25)
Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length $$\ell $$ and mass m. The rod is pivoted at its centre 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is :
_12th_January_Morning_Slot_en_25_1.png)
$${1 \over {2\pi }}\sqrt {{{3k} \over m}} $$
$${1 \over {2\pi }}\sqrt {{{6k} \over m}} $$
$${1 \over {2\pi }}\sqrt {{k \over m}} $$
$${1 \over {2\pi }}\sqrt {{{2k} \over m}} $$
Explanation
_12th_January_Morning_Slot_en_25_2.png)
$$\tau = - 2kx{\ell \over 2}\cos \theta $$
$$ \Rightarrow $$ $$\tau = \left( {{{K{\ell ^2}} \over 2}} \right)\theta = - C\theta $$
$$ \Rightarrow $$ $$f = {1 \over {2\pi }}\sqrt {{C \over 1}} = {1 \over {2\pi }}\sqrt {{{{{K{\ell ^2}} \over 2}} \over {{{M{\ell ^2}} \over {12}}}}} $$
$$ \Rightarrow $$ $$f = {1 \over {2\pi }}\sqrt {{{6K} \over M}} $$
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