Question

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is `x/2` times its original time period. Then the value of x is ______.

Options

• `2 sqrt 3`

• 4

• `sqrt 3`

• `sqrt 2`

Answer 1

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is `x/2` times its original time period. Then the value of x is `bbunderline(sqrt 2)`.

Explanation:

The time period T of a simple pendulum is given by the formula:

T = `2 pi sqrt(L/g)`

If the length is made half its original length, the new length is:

L' = `L/2`

The new time period is:

T' = `2 pi sqrt((L//2)/g)`

= `2 pi sqrt(L/(2 g)`

= `1/sqrt 2 (2 pi sqrt(L/g))`

= `1/sqrt T`

The problem states that the new time period of oscillation is `x/2` times its original time period:

T' = `x/2 T`

⇒ `1/sqrt 2 T = x/2 T`

⇒ `1/sqrt 2 = x/2`

⇒ x = `2/sqrt 2`

⇒ x = `sqrt 2`