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

Question

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 (A) 4 (B) 3 (C) 6 (D) 2

Tardigrade Admin · Accepted Answer

If t is the time taken by pendulums to come in same phase again first time after t=0. Then, (n-1) 2 π √(l1/g) =n 2 π √(l2/g) (n-1) × √1.44 =n (n-1) =1.2 n (n+1) =1.2 n n =(1/0.2)=5

Q. 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

3

6

2

5

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

Q. 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

4

3

6

2

5

If t is the time taken by pendulums to come in same phase again first time after t=0. Then, (n−1)2πgl1​​​=n2πgl2​​​ (n−1)×1.44​=n (n−1)=1.2n (n+1)=1.2n n=0.21​=5

Q. 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

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