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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 ______.
рдкрд░реНрдпрд╛рдп
`2 sqrt 3`
4
`sqrt 3`
`sqrt 2`
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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`
