JEE MAIN - Physics (2021 - 27th August Evening Shift - No. 18)
Two simple harmonic motion, are represented by the equations $${y_1} = 10\sin \left( {3\pi t + {\pi \over 3}} \right)$$ $${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$$ Ratio of amplitude of y1 to y2 = x : 1. The value of x is ______________.
Answer
1
Explanation
$${y_1} = 10\sin \left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ Amplitude = 10
$${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$$
$${y_2} = 10\left( {{1 \over 2}\sin 3\pi t + {{\sqrt 3 } \over 2}\cos 3\pi t} \right)$$
$${y_2} = 10\left( {\cos {\pi \over 3}\sin 3\pi t + \sin {\pi \over 3}\cos 3\pi t} \right)$$
$${y_2} = 10\left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ Amplitude = 10
So ratio of amplitudes = $${{10} \over {10}}$$ = 1
$${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$$
$${y_2} = 10\left( {{1 \over 2}\sin 3\pi t + {{\sqrt 3 } \over 2}\cos 3\pi t} \right)$$
$${y_2} = 10\left( {\cos {\pi \over 3}\sin 3\pi t + \sin {\pi \over 3}\cos 3\pi t} \right)$$
$${y_2} = 10\left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ Amplitude = 10
So ratio of amplitudes = $${{10} \over {10}}$$ = 1
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