Physics · 2019
JEE Main · 12 January 2019, Shift 2 · Q12
A simple harmonic motion is represented by: y=5( sin 3 π t+√ 3 cos 3 π t) cm The amplitude and time period of the motion are:
A simple harmonic motion is represented by:
$$y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) \mathrm{cm}
$$
The amplitude and time period of the motion are:
Official answer
(4)
$\displaystyle 10 \mathrm{~cm}, \frac{2}{3} \mathrm{~s}$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.