JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 23)
Two simple harmonic motions are represented by the equations
$${x_1} = 5\sin \left( {2\pi t + {\pi \over 4}} \right)$$ and $${x_2} = 5\sqrt 2 (\sin 2\pi t + \cos 2\pi t)$$. The amplitude of second motion is ................ times the amplitude in first motion.
$${x_1} = 5\sin \left( {2\pi t + {\pi \over 4}} \right)$$ and $${x_2} = 5\sqrt 2 (\sin 2\pi t + \cos 2\pi t)$$. The amplitude of second motion is ................ times the amplitude in first motion.
Answer
2
Explanation
$${x_2} = 5\sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2\pi t + {1 \over {\sqrt 2 }}\cos 2\pi t} \right)\sqrt 2 $$
$$ = 10\sin \left( {2\pi t + {\pi \over 4}} \right)$$
$$\therefore$$ $${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$$
$$ = 10\sin \left( {2\pi t + {\pi \over 4}} \right)$$
$$\therefore$$ $${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$$
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