A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be __________ s. [take g=π² m s⁻²]

Solution

Physics sir · 3 steps

At a height equal to R you are 2R from the centre, so g drops to a quarter. Use that g in the period.

1. Find the new g. Gravity goes as 1/r², where r is the distance from the earth's centre. At height h = R, that distance is r = 2R, so g' = g(R/2R)² = g/4 = π²/4 m s⁻².

   T = 2π√(L/g')

2. Put in numbers. L/g' = 4/(π²/4) = 16/π², so √(L/g') = 4/π. Then T = 2π × 4/π = 8 s.
3. So. The time period of small oscillations is 8 s.

The common slip: 4. Using the surface value g = π² instead of g/4 gives T = 2π × 2/π = 4 s, not 8.

The idea to keep. Height above the surface is not distance from the centre. Add R first, then use g' = g(R/(R+h))². A weaker g always means a slower pendulum.

Answer from the NTA final answer key. Solution by PrepWiser.