Two pendulums of lengths $1.44 m$ and $1 m$ start to swing together. The number of vibrations after which they will again start swinging together is

4

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3

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6

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2

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5

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Solution

The correct option is E $5$
If $t$ is the time taken by pendulums to come in same phase again first time after $t = 0$ .
Then, $(n + 1) 2 π \sqrt{\frac{l_{1}}{g}} = n 2 π \sqrt{\frac{l_{2}}{g}}$
$(n - 1) \times \sqrt{1.44} = n$
$(n - 1) = 1.2 n$
$(n + 1) = 1.2 n$
$n = \frac{1}{0.2} = 5$

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Two pendulums of lengths 1.44m and 1m start to swing together. The number of vibrations after which they will again start swinging together is

The correct option is E 5If t is the time taken by pendulums to come in same phase again first time after t=0.Then, (n+1)2π√l1g=n2π√l2g(n−1)×√1.44=n(n−1)=1.2n(n+1)=1.2nn=10.2=5

Two pendulums of lengths $1.44 m$ and $1 m$ start to swing together. The number of vibrations after which they will again start swinging together is

The correct option is E $5$
If $t$ is the time taken by pendulums to come in same phase again first time after $t = 0$ .
Then, $(n + 1) 2 π \sqrt{\frac{l_{1}}{g}} = n 2 π \sqrt{\frac{l_{2}}{g}}$
$(n - 1) \times \sqrt{1.44} = n$
$(n - 1) = 1.2 n$
$(n + 1) = 1.2 n$
$n = \frac{1}{0.2} = 5$