Question

Time period $T$ of a simple pendulum of length $l$ is given by $T=2\pi \sqrt{\frac{l}{g}}$. If the length is increased by $2\%$, then an approximate change in the time period is

• A. $2\%$
• B. $1\%$
• C. $\frac{1}{2}\%$
• D. None of these

Answer 1

The correct option is B $1\%$

$T=2\pi \sqrt{\frac{l}{g}}$
$T+\delta T=2\pi \sqrt{\frac{l(1+0.02)}{g}}$
Therefore
$\delta T=(T+\delta T)-T$
$=2\pi \sqrt{\frac{l(1+0.02)}{g}}-2\pi \sqrt{\frac{l}{g}}$
$=2\pi \sqrt{\frac{l}{g}}(\sqrt{1+0.02}-\sqrt{1})$
$=2\pi \sqrt{\frac{l}{g}}\left(\right(1+0.02{)}^{\frac{1}{2}}-1)$
$=2\pi \sqrt{\frac{l}{g}}((1+\frac{0.02}{2})-1)$ ...by applying binomial expansion and neglecting higher order terms
$=2\pi \sqrt{\frac{l}{g}}\left(0.01\right)$
$=\left(0.01\right)T$
Therefore $\%$ change is
$\frac{T+\delta T-\left(T\right)}{T}\times 100=\frac{\delta T}{T}\times 100$
$=\frac{0.01T}{T}\times 100=1\%$