Question

A simple pendulum has a time period T1, when on the earth’s surface; and T2, when taken to a height R above the earth’s surface (R is the radius of the earth). The value of T2/T1 is?

(a) 1

(b) √2

(c) 4

(d) 1

The question was posed to me during an internship interview.

I need to ask this question from Harmonic Motion in section Oscillations of Engineering Physics I

Answer 1

Correct answer is (d) 1

The explanation: g^‘/g=(R/(R+R))^2=1/4

As

T∝1/√g

T2/T1 = √(g/g^‘)=√(4/1)=2.