Question

Two pendulums at rest start swinging together. Their lengths are respectively 1.44 m and 1 m. They will
again start swinging in same phase together after (of longer pendulum)​

Answer 1

Answer

given,

length of two string (l₂) = 1 m

length of two string (l₁) =  1.44 m

initial phase after some time P

each pendulum has completed an integer number of cycles.

The longer pendulum has completed n  =P/T₁ oscillations

The shorter pendulum has completed m = P/T₂ oscillations

$\dfrac{T_1}{T_2}=\sqrt{\dfrac{l_1}{l_2}}$

$\dfrac{T_1}{T_2}=\sqrt{\dfrac{1.44}{1}}$

$\dfrac{T_1}{T_2}=1.2$

$\dfrac{T_1}{T_2}=\dfrac{12}{10}$

$\dfrac{T_1}{T_2}=\dfrac{6}{5}$

So the earliest time(smallest valuer of P) is given by m = 6 and n = 5.

Answer 2

Explanation:

given,

length of two string (l₂) = 1 m

length of two string (l₁) = 1.44 m

initial phase after some time P

each pendulum has completed an integer number of cycles.

The longer pendulum has completed n =P/T₁ oscillations

The shorter pendulum has completed m = P/T₂ oscillations

\dfrac{T_1}{T_2}=\sqrt{\dfrac{l_1}{l_2}}

T

2

T

1

=

l

2

l

1

\dfrac{T_1}{T_2}=\sqrt{\dfrac{1.44}{1}}

T

2

T

1

=

1

1.44

\dfrac{T_1}{T_2}=1.2

T

2

T

1

=1.2

\dfrac{T_1}{T_2}=\dfrac{12}{10}

T

2

T

1

=

10

12

\dfrac{T_1}{T_2}=\dfrac{6}{5}

T

2

T

1

=

5

6

So the earliest time(smallest valuer of P) is given by m = 6 and n = 5.