JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 14)
The point A moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30$$^\circ$$ in 0.1 s. The perpendicular projection 'P' from 'A' on the diameter MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be :
_25th_February_Evening_Shift_en_14_1.png)
9.87 N
0.49 N
50 N
100 N
Explanation
_25th_February_Evening_Shift_en_14_2.png)
The point a covers 30$$^\circ$$ in 0.1 sec.
Means $${\pi \over 6}\buildrel {} \over \longrightarrow 0.1$$ sec.
$$1\buildrel {} \over \longrightarrow {{0.1} \over {{\pi \over 6}}}$$
$$2\pi \buildrel {} \over \longrightarrow {{0.1 \times 6} \over \pi } \times 2\pi $$
$$T = 1.2$$ sec.
We know that $$\omega = {{2\pi } \over T}$$
$$\omega = {{2\pi } \over {1.2}}$$
Restoration force $$(F) = m{\omega ^2}A$$
Then Restoration force per unit mass $$\left( {{F \over m}} \right) = {\omega ^2}A$$
$$\left( {{F \over m}} \right) = {\left( {{{2\pi } \over {1.2}}} \right)^2} \times 0.36$$
$$ \cong 9.87$$ N
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