Oscillations & Simple Harmonic Motion
Stage 2 – Page 3 | Numerical Problems (Board Pattern)
Problem 1: Time Period of a Spring–Mass System
A mass of 2 kg is attached to a spring of force constant 800 N/m. Find the time period of oscillation.
Solution:
Given:
m = 2 kg
k = 800 N/m
Formula:
T = 2π√(m/k)
T = 2π√(2 / 800)
T = 2π√(1 / 400)
T = 2π × 1/20
T = π/10 s
Problem 2: Time Period of a Simple Pendulum
Find the time period of a pendulum of length 1 m at a place where g = 9.8 m/s².
Solution:
Formula:
T = 2π√(L/g)
T = 2π√(1 / 9.8)
T ≈ 2.01 s
Problem 3: Maximum Velocity in SHM
A particle executes SHM with amplitude 0.05 m and angular frequency 10 rad/s. Find maximum velocity.
Solution:
Formula:
vmax = ωA
vmax = 10 × 0.05
vmax = 0.5 m/s
Problem 4: Maximum Acceleration in SHM
Using the data from Problem 3, find maximum acceleration.
Solution:
Formula:
amax = ω²A
amax = (10)² × 0.05
amax = 5 m/s²
Problem 5: Displacement at Given Time
A particle executes SHM according to:
x = 0.1 sin(5t)
Find displacement at t = π/10 s.
Solution:
x = 0.1 sin(5 × π/10)
x = 0.1 sin(π/2)
x = 0.1 m
Problem 6: Energy of SHM
Find total energy of a particle of mass 1 kg executing SHM with ω = 10 rad/s and A = 0.1 m.
Solution:
Formula:
E = ½mω²A²
E = ½ × 1 × 100 × (0.1)²
E = 0.5 J
Problem 7: Change in Time Period
If the mass attached to a spring is doubled, how does time period change?
Solution:
T ∝ √m
If m becomes 2m:
T' = √2 T
Problem 8: Frequency of SHM
If the time period of an oscillator is 0.5 s, find its frequency.
Solution:
f = 1/T = 2 Hz
Board Exam Tips for Numericals
- Always write formula first
- Substitute values clearly
- Write final answer with unit
- Do not skip steps
Stage 2 – Page 3 Completed Successfully ✅
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🔹 Simple Harmonic Motion (SHM) — Core Series
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🔹 Simple Harmonic Motion (SHM) — Extended Series (30–56)
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