The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$, where $g$…

The time period $T$ of a simple pendulum of length $l$ is given by $T=2 \pi \sqrt{\frac{l}{g}}$, where $g$ denotes the acceleration due to gravity. If the length of the pendulum is increased by $1 \%$, then the approximate change in its time period is
  1. $0.5 \%$
  2. $2 \%$
  3. $1 \%$
  4. $4 \%$

Solution

Given, $T=2 \pi \sqrt{\frac{l}{g}}$ Then, $\frac{d T}{d l}=\frac{2 \pi}{\sqrt{g}} \cdot \frac{1}{2 \sqrt{l}}=\frac{T}{\sqrt{l}} \cdot \frac{1}{2 \sqrt{l}} \quad\left[\because T=\frac{2 \pi}{\sqrt{g}} \cdot \sqrt{l}\right]$ $ \begin{aligned} \frac{d T}{d l}=\frac{T}{2 l} & \Rightarrow \frac{d T}{T}=\frac{1}{2} \frac{d l}{l} \\ \text { Given, } \frac{d l}{l} \times 100 & =1 \Rightarrow \frac{d l}{l}=\frac{1}{100} \\ \Rightarrow \quad \frac{d T}{T} & =\frac{1}{2} \cdot \frac{1}{100} \\ \Rightarrow \quad \frac{d T}{T} \times 100 & =\frac{1}{2} \cdot \frac{1}{100} \times 100=\frac{1}{2}=0.5 \% \end{aligned} $ $\therefore$ Change in its time period $T$ is $0.5 \%$

Asked in: AP EAMCET 2021 (23 Aug Shift 1)

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