A simple pendulum has a length of 0.98 meters on Earth where g = 9.8 m/s^2. What is its period of small-angle oscillation T?

Answer and explanation

Correct answer: D. 1.99 seconds (approx 2.0 s)

  1. Step 1: The period of a simple pendulum for small amplitudes is: T = 2 pi sqrt(L / g).
  2. Step 2: Substitute L = 0.98 m and g = 9.8 m/s^2: L / g = 0.98 / 9.8 = 0.10 s^2.
  3. Step 3: Compute square root: sqrt(0.10) approx 0.3162.
  4. Step 4: Multiply by 2 pi: T = 2 3.1416 * 0.3162 approx 1.987 s approx 2.0 seconds.
Answer
1.99 seconds (approx 2.0 s).
Why not the others: '1.00 second' forgets the factor of 2 in 2pi; '3.14 seconds' is pi sqrt(L/g) without multiplying by 2; '0.50 seconds' inverts the period.

Practise more Physics: Mechanics, Electricity, and Modern Physics.

Free questions with an explanation for every one, and full timed General Secondary Certificate (Tawjihiya Kuwait) mock tests.

All Physics: Mechanics, Electricity, and Modern Physics questionsTake a mock test

More Physics: Mechanics, Electricity, and Modern Physics questions