Question

The time of oscillation t of a pendulum is given by t =2π√l/g Determine the approximate percentage error in t when L has a positive error of 0.2% and g has a negative error 0.1%

Answer 1

$\bf{\underline{\underline{Given}}}$

$\mathsf{\bigstar\: Time\: of \: oscillation\: (t) = 2\pi \sqrt{\frac{L}{g}}}$

$\mathsf{\bigstar\: Error\: \% \: in\: L = 0.2\: \%}$

$\mathsf{\bigstar\: Error\: \% \: in\: g = - \: 0.1\: \%}$

$\bf{\underline{\underline{To\:find}}}$

$\mathsf{\bigstar\: Percentage\:\: error\:\: in \:\: t }$

$\bf{\underline{\underline{Solution}}}$

We know that

$\mathtt{If \: \: A = B^{m} × C^{n}}$

$\fbox{\mathtt{\red{\frac{\Delta A}{A} = m \frac{\Delta B}{B} + n \frac{\Delta C}{C}}}}$

Now,

$\mathtt{\implies\: t = 2\pi \sqrt{\frac{L}{g}}}$

$\mathtt{\implies\: t = 2\pi (\frac{L}{g})^{\frac{1}{2}}}$

$\mathtt{\implies\: t = 2\pi (\frac{L^{\frac{1}{2}}}{g^{\frac{1}{2}}})}$

$\mathtt{\implies\: t = 2\pi L^{\frac{1}{2}} \: g^{\frac{-1}{2}}}$

Finding fractional error in 't'

$\fbox{\mathtt{\orange{\frac{\Delta t}{t} = \frac{1}{2}\: (\frac{\Delta L}{L}) + \frac{- 1}{2} \: (\frac{\Delta g}{g})}}}$

$\mathtt{\implies\: \frac{\Delta t}{t} = \frac{1}{2} (0.2 \: \%) + \frac{-1}{2} (-0.1 \: \%)}$

$\mathtt{\implies\: \frac{\Delta t}{t} = \frac{1}{\cancel{2}} (\cancel{0.2} \: \%) + \frac{1}{\cancel{2}} (\cancel{0.1}\: \%)}$

$\mathtt{\implies\: \frac{\Delta t}{t} = 0.1\: \% + 0.05\: \%}$

$\mathtt{\implies\: \frac{\Delta t}{t} = 0.15\: \%}$

For percentage error, Just Multiply the fractional error by 100

$\mathtt{\implies\: 100 × \frac{\Delta t}{t} = 100 × 0.15\: \% }$

$\mathtt{\implies\: t \: \% = 15\: \%\: \: \green{(Answer)}}$

Therefore, t has percentage error of 15%

_____________________________

Note: Constant such as 2π has no error (either fractional or percentage)

Answer 2

Answer:

t has a percentage error of 15°

Step-by-step explanation:

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