Find Maximum Percentage Error Using g and Time Period

 

❓ Question

The percentage error in acceleration due to gravity $g$ is $4\%$, and the percentage error in time period $T$ is $3\%$.

For a simple pendulum, find the percentage error in energy $E$.

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✍️ Solution

For a simple pendulum,

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

Squaring,

$$T^2=4\pi^2\frac{l}{g}$$

Therefore,

$$l=\frac{T^2g}{4\pi^2}$$

The energy of a simple pendulum is

$$E=mg\,l\theta^2$$

Considering $m$ and $\theta$ as constants,

$$E\propto gl$$

Substituting,

$$l\propto T^2g$$

Therefore,

$$E\propto g(T^2g)$$
$$E\propto T^2g^2$$

For maximum percentage error,

$$\frac{\Delta E}{E}\times100 = 2\frac{\Delta T}{T}\times100 + 2\frac{\Delta g}{g}\times100$$

Substituting the values,

$$\%\text{ error in }E = 2(3)+2(4)$$
$$=6+8$$
$$=14\%$$

✅ Final Answer

$$\boxed{14\%}$$