The periodic time, [tex]$T$[/tex], of a pendulum varies as the square root of its length, [tex]$l$[/tex]. If [tex][tex]$T=6$[/tex][/tex] when [tex]$l=g$[/tex], find [tex]$T$[/tex] when [tex][tex]$l=25$[/tex][/tex].



Answer :

Certainly! Let's go through this problem step-by-step.

The periodic time [tex]\( T \)[/tex] of a pendulum is given to vary as the square root of its length [tex]\( l \)[/tex].

This means we can write:
[tex]\[ T \propto \sqrt{l} \][/tex]

To express this as an equation, we introduce a constant of proportionality [tex]\( k \)[/tex]:
[tex]\[ T = k \sqrt{l} \][/tex]

Given that:
[tex]\[ T = 6 \quad \text{when} \quad l = 9 \][/tex]

We can substitute these values into the equation to find [tex]\( k \)[/tex]:
[tex]\[ 6 = k \sqrt{9} \][/tex]

Since [tex]\( \sqrt{9} = 3 \)[/tex], we have:
[tex]\[ 6 = k \cdot 3 \][/tex]

Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{6}{3} = 2 \][/tex]

Now, we need to find [tex]\( T \)[/tex] when [tex]\( l = 25 \)[/tex]. Using the proportionality with the constant [tex]\( k \)[/tex] we just found:
[tex]\[ T = k \sqrt{25} \][/tex]

Since [tex]\( k = 2 \)[/tex] and [tex]\( \sqrt{25} = 5 \)[/tex]:
[tex]\[ T = 2 \cdot 5 = 10 \][/tex]

Therefore, the periodic time [tex]\( T \)[/tex] when [tex]\( l = 25 \)[/tex] is:
[tex]\[ T = 10 \][/tex]