JEE Advanced Previous Year Questions (2018 - 2024): Simple Harmonic Motion | Physics for JEE Main & Advanced PDF Download

2024

Q1: Two particles, 1 and 2, each of mass m, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x₀, are oscillating with amplitude a and angular frequency ω. Thus, their positions at time t are given by x₁(t) = (x₀ + d) + a sin ωt and x₂(t) = (x₀ - d) - a sin ωt, respectively, where d > 2a.
Particle 3 of mass m moves towards this system with speed u₀ = aω/2, and undergoes instantaneous elastic collision with particle 2, at time t₀. Finally, particles 1 and 2 acquire a center of mass speed v_cm and oscillate with amplitude b and the same angular frequency ω.

If the collision occurs at time t₀ = 0, the value of v_cm / (aω) will be _____.       [JEE Advanced 2024 Paper 2]
Ans: 0.75
At T t$0$₀ = 0
Before collision
After collision
v_CM = m ⋅ aω2 + m ⋅ aωm + m
v_CM = 3aω4
V_CMaω = 34

V_CMaω = 0.75

Q2: Two particles, 1 and 2, each of mass m, are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at x₀, are oscillating with amplitude a and angular frequency ω. Thus, their positions at time t are given by x₁(t) = (x₀ + d) + a sin ωt and x₂(t) = (x₀ - d) - a sin ωt, respectively, where d > 2a.
Particle 3 of mass m moves towards this system with speed u₀ = aω/2, and undergoes instantaneous elastic collision with particle 2, at time t₀. Finally, particles 1 and 2 acquire a center of mass speed v_cm and oscillate with amplitude b and the same angular frequency ω.

If the collision occurs at time t₀ = π/(2ω), then the value of 4b²/a² will be _____.     [JEE Advanced 2024 Paper 2]
Ans: 4.25
t₀ = _π2ω = T4
Particles are at extreme position
After collision
in C-frame
using WET,
W_spring = ΔK
12 k (2b)² - 12 k (2a)² = 2 × 12 m × (a4) ² (k = spring constant)
4kb² - 4ka² = 2 × m × a²16 × 2km
4b² = 174 a²
4b²a² = 4.25

2023

So, we can calculate the focal length :

Next, we need to calculate the error in the determination of the focal length. For that, we find the differential of the lens formula :
Then, the derivative of the equation gives us the change in the focal length (df) in terms of the changes in the object distance (du) and the image distance (dv): 

For maximum error, we get :

This equation tells us how errors in (u) and v propagate to an error in f. Now, when you compute the relative error in the focal length, you get :

Plugging in your values of  u = 10 cm,  du =0.1 cm,  v = 20 cm,  dv = 0.2 cm, and  f = 20/3 cm, you indeed get: 

So, the error in the focal length of the lens is indeed 1% (i.e., n = 1).

2022

Q1: In a particular system of units, a physical quantity can be expressed in terms of the electric charge e, electron mass m_e, Planck's constant ℎ, and Coulomb's constant , where ε₀ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is . The value of α + β + γ + δ is______. [JEE Advanced 2022 Paper 2]
Ans: 

Compare :

On Solving:

2019

∴ f = 20 cm Also, 

= 50/36
= 1.38 and 1.39 (both)

2018

We know that

where W is weight or load = mg = 1.2 × 10 = 12 kg m s⁻², Y is Young's modulus = 2 × 10¹¹ N m⁻², L is length of wire with load = 1.0 m, A is area of steel wire
=

Therefore,

Now, least count of vernier scale =

Therefore, Vernier reading =
Vernier reading = 0.3 mm / 0.1 mm = 3
Therefore, 3^rd vernier scale division coincides with the main scale division.