 ### JEE Mains Previous Years Questions with Solutions

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### JEE Main 2021 (Online) 31st August Morning Shift

Numerical
A particle of mass 1 kg is hanging from a spring of force constant 100 Nm$$-$$1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $${T \over x}$$. The value of x is _____________.

## Explanation KE = PE

$$y = {A \over {\sqrt 2 }} = A\sin \omega t$$ $$t = {T \over 8} = {T \over x}$$

x = 8
2

### JEE Main 2021 (Online) 27th August Evening Shift

Numerical
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 ______________.

## 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
3

### 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.

## Explanation

$${x_2} = 5\sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2\pi t + {1 \over {\sqrt 2 }}\cos 2\pi t} \right)\sqrt 2$$

$$= 10\sin \left( {2\pi t + {\pi \over 4}} \right)$$

$$\therefore$$ $${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$$
4

### JEE Main 2021 (Online) 27th July Evening Shift

Numerical
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 _________________.

## 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

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