📘 Numericals – Simple Pendulum
Q1. A simple pendulum completes 40 oscillations in 1 minute. Find frequency and time period.
Number of oscillations = 40, Time = 60 s
Frequency (f) = oscillations / time = 40/60 = 0.67 Hz
Time period (T) = 1/f = 1/0.67 ≈ 1.5 s
Q2. Time period of a pendulum is 2 s. Find its frequency.
Given T = 2 s
f = 1/T = 1/2 = 0.5 Hz
Q3. A seconds pendulum is taken to a place where g becomes one-fourth. Find new time period.
T ∝ 1/√g
If g becomes g/4 → √g becomes √(g/4) = √g/2
New T = 2 × old T = 2 × 2 = 4 s
Q4. Find length of seconds pendulum (g = 10 m/s²).
Formula: T = 2π√(l/g)
For seconds pendulum, T = 2 s
l = gT² / (4π²) = 10×4 / (4×3.14²)
l ≈ 1.01 m
Q5. Compare time periods of pendulums of lengths 1 m and 9 m.
T ∝ √l
T₁/T₂ = √(1/9) = 1/3
Ratio = 1 : 3
Q6. A pendulum completes 6 oscillations in 5 s. Find time period and length (g = 9.8 m/s²).
T = total time / oscillations = 5/6 = 0.83 s
l = gT² / (4π²)
l ≈ 0.17 m (approx)
Q7. Time periods ratio = 2 : 1. Find lengths ratio.
T ∝ √l → l ∝ T²
l₁/l₂ = (2/1)² = 4/1
Ratio = 4 : 1
Q8. Time from mean to extreme = 0.2 s. Find time period.
Mean to extreme = T/4
T = 4 × 0.2 = 0.8 s
Q9. Time from one extreme to other = ?
Extreme to extreme = T/2
Time = 1 s (for seconds pendulum)
