JEE Main SHM PYQ

1.(JEE Main 2021 (Online) 18th March Evening Shift)

The function of time representing a simple harmonic motion with a period of $\frac{π}{ω}$ is :

A. cos( $ω$ t) + cos(2 $ω$ t) + cos(3 $ω$ t)jx-math class="MJX-TEX" aria-hidden="true">//--> $ω$ t)

B. sin²( $ω$ t)

C. sin( $ω$ t) + cos( $ω$ t)

D. 3cos $(\frac{π}{4} - 2 ω t)$

Correct Answer is Option ()

2.(JEE Main 2021 (Online) 27th August Evening Shift )

Two simple harmonic motion, are represented by the equations $y_{1} = 10 \sin (3 π t + \frac{π}{3})$ $y_{2} = 5 (\sin 3 π t + \sqrt{3} \cos 3 π t)$ Ratio of amplitude of y₁ to y₂ = x : 1. The value of x is ______________.

Correct Answer is (1)

$y_{1} = 10 \sin (3 π t + \frac{π}{3})$ $\Rightarrow$ Amplitude = 10

$y_{2} = 5 (\sin 3 π t + \sqrt{3} \cos 3 π t)$

$y_{2} = 10 (\frac{1}{2} \sin 3 π t + \frac{\sqrt{3}}{2} \cos 3 π t)$

$y_{2} = 10 (\cos \frac{π}{3} \sin 3 π t + \sin \frac{π}{3} \cos 3 π t)$

$y_{2} = 10 (3 π t + \frac{π}{3})$ $\Rightarrow$ Amplitude = 10

So ratio of amplitudes = $\frac{10}{10}$ = 1

3.(JEE Main 2021 (Online) 26th August Evening Shift )

Two simple harmonic motions are represented by the equations

$x_{1} = 5 \sin (2 π t + \frac{π}{4})$ and $x_{2} = 5 \sqrt{2} (\sin 2 π t + \cos 2 π t)$ . The amplitude of second motion is ................ times the amplitude in first motion.

Correct Answer is (2)

$x_{2} = 5 \sqrt{2} (\frac{1}{\sqrt{2}} \sin 2 π t + \frac{1}{\sqrt{2}} \cos 2 π t) \sqrt{2}$

$= 10 \sin (2 π t + \frac{π}{4})$

$∴$ $\frac{A_{2}}{A_{1}} = \frac{10}{5} = 2$

4.(JEE Main 2019 (Online) 12th January Evening Slot)

A simple harmonic motion is represented by :

y = 5 (sin 3 $π$ t + $\sqrt{3}$ cos 3 $π$ t) cm

The amplitude and time period of the motion are :

A. 10 cm, $\frac{3}{2}$ s

B. 5 cm, $\frac{2}{3}$ s

C. 5 cm, $\frac{3}{2}$ s

D. 10D cm, $\frac{2}{3}$ s

Correct Answer is Option (D)

JEE Main 2019 (Online) 12th January Evening Slot Physics - Simple Harmonic Motion Question 86 English Explanation

y = 5[sin(3 $π$ t) + $\sqrt{3}$ cos(3 $π$ t)]

= 10sin $(3 π t + \frac{π}{3})$

Amplitude = 10 cm

T = $\frac{2 π}{w}$ = $\frac{2 π}{3 π}$ = $\frac{2}{3}$ sec

5.(JEE Main 2018 (Online) 15th April Evening Slot)

Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures.
x(t) = A sin (at + $δ$ )
y(t) = B sin (bt)

Identify the correct match below.

A. Parameters A $\neq$ B, a = b; $δ$ = 0;
Curve Parabola

B. Parameters A = B, a = b; $δ$ = $π / 2$
Curve Line

C. Parameters A $\neq$ B, a = b; $δ$ = $π / 2$
Curve Ellipse

D. Parameters A = B, a = 2b; $δ$ = $π / 2$
Curve Circle

Correct Answer is Option (C)

The given simple harmonic motions to form Lissajous figures are $x (t) = A \sin (a t + δ)$ and $y (t) = B \sin (b t)$ .

For parabola, conditions should be

A = B or A $\neq$ B, a = 2b, $δ$ = $π$ /2

For line, conditions should be

A = B, a = b, $δ$ = $π$

For circle, condition should be

A = B, a = b; $δ$ = $π$ /2

For ellipse, condition should be

A $\neq$ B, a = b; $δ$ = $π$ /2

Therefore, we obtain an ellipse.