Question

A simple pendulum of length 'L' is suspended from a roof of a trolley. A trolley moves in horizontal direction with an acceleration 'a'. What would be the period of oscillation of a simple pendulum?

(g is acceleration due to gravity)

विकल्प

• `2pisqrt("L"/(g+a))`

• `2pisqrt"L"(a^2+g^2)^(1/4)`

• `2pisqrt("L"/(g-a))`

• `2pisqrt"L"(a^2+g^2)^(1/2)`

Answer 1

`2pisqrt"L"(a^2+g^2)^(1/4)`

Explanation:

`T=2pisqrt(L/g)`

where g is the effective value of acceleration due to gravity. When the trolley has acceleration 'a' in horizontal direction, the effective value of acceleration is the resultant of a and g which 1 are at right angles to each other. Hence effective acceleration is `sqrt(a^2+g^2)` or `(a^2+g^2)^(1/2)`

`thereforeT=2pisqrt(L/(a^2+g^2)^(1/2))=(2pisqrtL)/(sqrt((a^2+g^2)^(1/2)))`

`=2pisqrtL(a^2+g^2)^(-1/4)`