JEE MAIN - Physics (2023 - 29th January Evening Shift - No. 21)
A particle of mass 250 g executes a simple harmonic motion under a periodic force $$\mathrm{F}=(-25~x)\mathrm{N}$$. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.
Answer
40
Explanation
$$F = - 25x$$
$$.250{{{d^2}x} \over {d{t^2}}} = - 25x$$
$${{{d^2}x} \over {d{t^2}}} = - 100x$$
$$ \Rightarrow \omega = 10$$ rad/sec
& $$\omega A = {v_{\max }}$$
$$10\,A = 4$$
$$ \Rightarrow A = 0.4$$ m
$$ = 40$$ cm
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