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 ______________.
So ratio of amplitudes = $${{10} \over {10}}$$ = 1
2
JEE Main 2021 (Online) 26th August Evening Shift
Numerical
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.
A particle executes simple harmonic motion represented by displacement function as
x(t) = A sin($$\omega$$t + $$\phi$$)
If the position and velocity of the particle at t = 0 s are 2 cm and 2$$\omega$$ cm s$$-$$1 respectively, then its amplitude is $$x\sqrt 2 $$ cm where the value of x is _________________.
Your Input ________
Answer
Correct Answer is 2
Explanation
x(t) = A sin($$\omega$$t + $$\phi$$)
v(t) = A$$\omega$$ cos ($$\omega$$t + $$\phi$$)
2 = A sin$$\phi$$ ...... (1)
2$$\omega$$ = A$$\omega$$ cos$$\phi$$ ....... (2)
From (1) and (2)
tan$$\phi$$ = 1
$$\phi$$ = 45$$^\circ$$
Putting value of $$\phi$$ in equation (1),
$$2 = A\left\{ {{1 \over {\sqrt 2 }}} \right\}$$
$$A = 2\sqrt 2 $$
$$ \therefore $$ x = 2
4
JEE Main 2021 (Online) 25th July Morning Shift
Numerical
A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60$$^\circ$$ to the vertical will be (g = 10 m/s2) ____________ m/s.