JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 19)
A simple harmonic motion is represented by :
y = 5 (sin 3 $$\pi $$ t + $$\sqrt 3 $$ cos 3 $$\pi $$t) cm
The amplitude and time period of the motion are :
y = 5 (sin 3 $$\pi $$ t + $$\sqrt 3 $$ cos 3 $$\pi $$t) cm
The amplitude and time period of the motion are :
10 cm, $${3 \over 2}$$ s
5 cm, $${2 \over 3}$$ s
5 cm, $${3 \over 2}$$ s
10 cm, $${2 \over 3}$$ s
Explanation
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y = 5[sin(3$$\pi $$t) + $$\sqrt 3 $$cos(3$$\pi $$t)]
= 10sin $$\left( {3\pi t + {\pi \over 3}} \right)$$
Amplitude = 10 cm
T = $${{2\pi } \over w}$$ = $${{2\pi } \over {3\pi }}$$ = $${2 \over 3}$$ sec
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