5. If a simple pendulum with mass m, length L has a time-period T on Earth, its time-period when
moved to a planet with one-fourth the acceleration due to gravity will be
a. 2*T
b. 4*T
c. T/2
d. T/4
✅ Correct Answer: (a) 2T
On a planet where gravity is one-fourth of Earth’s, the time period of the same simple pendulum becomes twice its value on Earth.
Concept: Time Period of a Simple Pendulum
For small oscillations, the time period T of a simple pendulum of length L in a gravitational field of acceleration g is:
T = 2π√(L/g)
This shows:
- T ∝ √L (directly proportional to square root of length)
- T ∝ 1/√g (inversely proportional to square root of gravitational acceleration)
- Mass m of the bob does not affect the time period, provided oscillations are small.
Step-by-Step Solution
Given:
- On Earth: gravity = g, period = T
- On another planet: gravity g’ = g/4
New time period T’ on the planet:
T' = 2π√(L/g') = 2π√(L/(g/4)) = 2π√(4L/g) = 2π·2√(L/g) = 2T
So the pendulum’s time period becomes twice the original when gravity becomes one-fourth, hence option (a) 2T is correct.
Option-by-Option Explanation
✅ Option (a) 2T – Correct
Since T ∝ 1/√g, reducing g by a factor of 4 multiplies T by √4 = 2.
Therefore, the new time period is T’ = 2T, matching option (a).
❌ Option (b) 4T – Incorrect
To get T’ = 4T, gravity would need to be reduced by a factor of 4² = 16, because T scales with 1/√g.
Here g is only reduced to one-fourth, not one-sixteenth, so 4T is an overestimate.
❌ Option (c) T/2 – Incorrect
T/2 would mean the pendulum oscillates faster (shorter period), which happens if gravity increases, not decreases.
Since g is smaller on the new planet, the pendulum must swing more slowly, giving a larger period, not T/2.
❌ Option (d) T/4 – Incorrect
T/4 would require gravity to increase by a factor of 4² = 16, making oscillations much faster.
Here gravity is weaker, so the time period cannot become a quarter of its original value.
