Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

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 _____________.

Your Input ________

Correct Answer is **8**

KE = PE

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

$$t = {T \over 8} = {T \over x}$$

x = 8

2

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 y_{1} to y_{2} = x : 1. The value of x is ______________.

Your Input ________

Correct Answer is **1**

$${y_1} = 10\sin \left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ 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

3

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.

$${x_1} = 5\sin \left( {2\pi t + {\pi \over 4}} \right)$$ 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.

Your Input ________

Correct Answer is **2**

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

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

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

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

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

4

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 _________________.

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

Your Input ________

Correct Answer is **2**

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

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

On those following papers in Numerical

Number in Brackets after Paper Indicates No. of Questions

JEE Main 2021 (Online) 31st August Morning Shift (1)

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