# Question 1ad48

Jan 19, 2018

Time period will decrease by 6 %

#### Explanation:

Suppose,the initial length of the pendulum was $100 l$ and time period was $T 1$

So,we can write,
$T 1 = 2 \pi \sqrt{100 \frac{l}{g}}$

If,length is decreased by 10 %, new length becomes $90 l \mathmr{and} \mathmr{if} t i m e p e r i o d i s n o w$T2 then we can write,

$T 2 = 2 \pi \sqrt{90 \frac{l}{g}}$

Dividing $T 2$ by $T 1$ we get,
$\frac{T 2}{T 1} = \sqrt{\frac{90}{100}}$
Or, $\frac{T 2}{T 1} = 0.94$
Or, $1 - \frac{T 2}{T 1} = 1 - 0.94$
Or, $\frac{T 1 - T 2}{T 1} = 0.06$
So, $\frac{\partial T}{T 1} \cdot 100 = 6$

Jan 19, 2018

See the explanation below

#### Explanation:

The period of a pendulum of $\text{length} = l$ is

$T = 2 \pi \sqrt{\frac{l}{g}}$

The new length of the pendulum is

$L = 0.9 l$

The new period is

${T}_{N} = 2 \pi \sqrt{\frac{0.9 l}{g}}$

${T}_{N} = \left(\sqrt{0.9}\right) 2 \pi \sqrt{\frac{l}{g}}$

${T}_{N} = \sqrt{0.9} T$

The new period is ${T}_{N} = 0.95 T$