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⏱ Chapter 09 · Practice Section

Practice with Timer & Score

Three difficulty levels. Select, time yourself, attempt, then reveal. Track your score honestly — the goal is improvement, not just completion.

Practice Timer
00:00

📊 Session Score

0
Correct
0
Wrong
0
Attempted
🎯 CBSE Level Strategy

Target: complete each question in 2–3 minutes. All answers can be found directly from formulas. Focus on showing proper steps in boards format.

EasyCBSE 2022
Q1 of 5

A particle executes SHM with time period T = 2π seconds. What is its angular frequency?

A
2π rad/s
B
1 rad/s
C
π rad/s
D
2 rad/s
EasyCBSE Formula
Q2 of 5

A simple pendulum has time period T on Earth. On the Moon (g_moon = g_earth/6), what is the new time period?

A
T/6
B
T/√6
C
T√6
D
6T
EasyCBSE Standard
Q3 of 5

In SHM, when is potential energy maximum?

A
At equilibrium position
B
At mean position
C
At extreme positions (x = ±A)
D
At x = A/2
Easy
Q4 of 5

A spring of constant k = 400 N/m is attached to a 1 kg mass. The time period of oscillation is:

A
2π/10 s
B
π/10 s
C
2π s
D
π/5 s
EasyCBSE 2023
Q5 of 5

The phase difference between displacement and acceleration in SHM is:

A
0
B
π/2
C
π
D
3π/2
🎯 NEET / JEE Main Level Strategy

Target: 90 seconds per question (NEET pace). For JEE Main MCQ: 2 minutes. For JEE Main numerical: 3 minutes. Don't rush — one wrong MCQ = −1 mark in JEE.

ModerateNEET 2020
Q6 of 10

Which of the following functions represents simple harmonic motion?

A
x = sin(ωt) + cos(2ωt)
B
x = sin(ωt) + cos(ωt)
C
x = sin(ωt) · cos(ωt)
D
x = sin²(ωt)
ModerateJEE Main 2022
Q7 of 10

A spring of spring constant k is cut into 3 equal parts. These 3 parts are connected in parallel. The effective spring constant and new time period for mass m are:

A
k_eff = k/3, T = 2π√(3m/k)
B
k_eff = 3k, T = 2π√(m/3k)
C
k_eff = 9k, T = 2π√(m/9k) = T₀/3
D
k_eff = 9k, T = 2π√(m/k)
ModerateNEET 2019
Q8 of 10

In SHM, the ratio of maximum velocity to maximum acceleration is:

A
Aω²
B
1/ω
C
ω
D
A/ω
ModerateJEE Main Numerical
Q9 of 10

A 2 kg block attached to a spring (k = 200 N/m) performs SHM. At x = 0.05 m from equilibrium, its velocity is 0.3 m/s. The amplitude of SHM is _____ × 10⁻² m.

ModerateNEET 2021
Q10 of 10

A pendulum of length L₁ has time period T₁. Another pendulum of length L₂ = 4L₁ has time period T₂. The ratio T₂/T₁ is:

A
4
B
2
C
1/2
D
1/4
🎯 JEE Advanced Level Strategy

Target: 4–5 minutes per question. These require multi-step analysis. Read the ENTIRE question before attempting. Identify all given information. Work systematically — one step error destroys everything.

AdvancedJEE Advanced StyleMulti-Correct
Q11 · Multiple Correct

A particle of mass 1 kg executes SHM: x = 0.1 sin(10πt + π/6) m. Select all CORRECT statements:

A
Amplitude = 0.1 m
B
Time period = 10 s
C
Maximum acceleration = 10π² m/s²
D
At t = 0, the particle is at x = 0.05 m moving in +x direction
AdvancedJEE Advanced 2021 Style
Q12 · Paragraph Based

PASSAGE: Two SHMs are: x₁ = 3 sin(ωt) cm and x₂ = 4 cos(ωt) cm. A particle undergoes the resultant of these two SHMs simultaneously.

The resultant amplitude of the superposition is:

A
3 cm
B
4 cm
C
5 cm
D
7 cm
AdvancedJEE Advanced
Q13 · Analytical

A particle moves under potential U(x) = 2x² − 4x. Find the angular frequency of small oscillations about the equilibrium position.

A
ω = 1 rad/s (if m = 1 kg)
B
ω = 2 rad/s (if m = 1 kg)
C
ω = 4 rad/s (if m = 1 kg)
D
ω = √2 rad/s (if m = 1 kg)
← Advanced Thinking Next: Exam Strategy →