JEE MAIN - Physics (2021 - 24th February Evening Shift - No. 1)
In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is :
_24th_February_Evening_Shift_en_1_1.png)
$${1 \over {2\pi }}\sqrt {{{2k} \over {Mg\sin \alpha }}} $$
$${1 \over {2\pi }}\sqrt {{k \over {Mg\sin \alpha }}} $$
$${1 \over {2\pi }}\sqrt {{{2k} \over M}} $$
$${1 \over {2\pi }}\sqrt {{k \over {2M}}} $$
Explanation
Let T be the time period of oscillation, then
$$T = 2\pi \sqrt {{M \over {{k_{eq}}}}} $$
$$\therefore$$ $$T = 2\pi \sqrt {{M \over {2k}}} $$ [$$\because$$ $${k_{eq}} = k + k$$]
_24th_February_Evening_Shift_en_1_2.png)
and frequency $$(f) = {1 \over T} = {1 \over {2\pi }}\sqrt {{{2k} \over M}} $$
$$T = 2\pi \sqrt {{M \over {{k_{eq}}}}} $$
$$\therefore$$ $$T = 2\pi \sqrt {{M \over {2k}}} $$ [$$\because$$ $${k_{eq}} = k + k$$]
and frequency $$(f) = {1 \over T} = {1 \over {2\pi }}\sqrt {{{2k} \over M}} $$
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