JEE Main SHM PYQ

21. (JEE Main 2021 (Online) 26th February Evening Shift )

Time period of a simple pendulum is T. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{α}{β} T$ . The value of $α$ is _________.

Correct Answer is (7)

JEE Main 2021 (Online) 26th February Evening Shift Physics - Simple Harmonic Motion Question 66 English Explanation
$\frac{5}{8}$ oscillation = $\frac{1}{2}$ oscillation + $\frac{1}{8}$ oscillation

From figure, A to B = $\frac{1}{2}$ oscillation and B to C is $\frac{1}{8}$ oscillation.

$∴$ $π + θ = ω t$

$\Rightarrow$ $π + \frac{π}{6} = ω t$

$\Rightarrow$ $\frac{7 π}{6} = (\frac{2 π}{T}) t$

$\Rightarrow$ t = $\frac{7 T}{12}$

22. (JEE Main 2019 (Online) 10th January Evening Slot)

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

A. $\frac{4 π}{3}$

B. $\frac{3}{8} π$

C. $\frac{7}{3} π$

D. $\frac{8 π}{3}$

Correct Answer is Option (D)

$v = ω \sqrt{A^{2} - x^{2}}$     . . .(1)

$a = - ω^{2} x$                . . .(2)

$| v | = | a |$                    . . .(3)

$ω \sqrt{A^{2} - x^{2}} = ω^{2} x$

$A^{2} - x^{2} = ω^{2} x^{2}$

$5^{2} - 4^{2} = ω^{2} (4^{2})$

$\Rightarrow 3 = ω \times 4$

$T = 2 π / ω$

23. (JEE Main 2018 (Online) 16th April Morning Slot)

A particle executes simple harmonic motion and is located at x = a, b and c at times t₀, 2t₀ and 3t₀ respectively. The freqquency of the oscillation is :

A. $\frac{1}{2 π t_{0}} \cos^{- 1} (\frac{a + c}{2 b})$

B. $\frac{1}{2 π t_{0}} \cos^{- 1} (\frac{a + b}{2 c})$

C. $\frac{1}{2 π t_{0}} \cos^{- 1} (\frac{2 a + 3 c}{b})$

D. $\frac{1}{2 π t_{0}} \cos^{- 1} (\frac{a + 2 b}{3 c})$

Correct Answer is Option ()

24 . (JEE Main 2017 (Online) 8th April Morning Slot)

The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s⁻¹ . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is $\frac{π}{4}$ .

A. 500 m/s²

B. 500 $\sqrt{2} m /$ s²

C. 750 m/s²

D. 750 $\sqrt{2}$ m / s²

Correct Answer is Option (B)

Mximum velocity, V_max = a $ω$

Maximum acceleration, A_max = a $ω$ ²

Given that,

$\frac{a ω^{2}}{a ω}$ = 10

$\Rightarrow$ $ω$ = 10 s^{$-$ 1}

Displacement, x = a sin ( $ω$ t + $\frac{π}{4}$ )

at t = 0, displacement x = 5

$∴$ 5 = a sin $(\frac{π}{4})$

$\Rightarrow$ 5 = a $\times$ $\frac{1}{\sqrt{2}}$

$\Rightarrow$ a = 5 $\sqrt{2}$

$∴$ Maximum acceleration,

A_max = a $ω$ ² = 5 $\sqrt{2}$ $\times$ (10)²

= 500 $\sqrt{2}$ m/s²

25.(JEE Main 2016 (Online) 9th April Morning Slot)

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and I, respectively. At time t = 0 one particle has displacement A while the other one has displacement $\frac{- A}{2}$ and they are moving towards each other. If they cross each other at time t, then t is :

A. $\frac{T}{6}$ le>//--> $\frac{T}{6}$

B. $\frac{5}{T}$

C. $\frac{T}{3}$

D. $\frac{T}{4}$

Correct Answer is Option (A)

JEE Main 2016 (Online) 9th April Morning Slot Physics - Simple Harmonic Motion Question 98 English Explanation
Angular displacement ( $θ$ ₁) of particle 1. from equilibrium,

$y_{1}$ = A sin $θ$ ₁

$\Rightarrow$    A = Asin $θ$ ₁

$\Rightarrow$    sin $θ$ ₁ = 1 = sin $\frac{π}{2}$

$∴$     $θ$ ₁ = $\frac{π}{2}$

Similarly for particle 2 angular displacement $θ$ ₂ from equilibrium,

y₂ = Asin $θ$ ₂

$\Rightarrow$     $-$ $\frac{A}{2}$ = Asin $θ$ ₂

$\Rightarrow$    sin $θ$ ₂ = $-$ $\frac{1}{2}$ = sin $(- \frac{π}{3})$

$\Rightarrow$     $θ$ ₂ = $-$ $\frac{π}{3}$

Relative angular displacement of the two particle,

$θ$ = $θ$ ₁ $-$ $θ$ ₂

= $\frac{π}{2}$ $-$ $(- \frac{π}{6})$

= $\frac{2 π}{3}$

Relative angular velocity $=$ $ω - (- ω)$ = $2 ω$

If they cross each other at time t

then,   t = $\frac{θ}{2 ω}$ = $\frac{2 π}{3 \times 2 ω}$ = $\frac{π}{3 \times \frac{2 π}{T}}$ = $\frac{T}{6}$