Question f3947

Sep 18, 2017

% $\text{error}$ $=$ 3%

Explanation:

The equation for the time period of a pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.

We need to evaluate the percentage error in the measurement of the time period $T$.

First, let's consider $\frac{L}{g}$.

These are two values being divided, so their percentage errors should be added together:

Rightarrow % "error of" $\frac{L}{g}$ = % "error of" L + % "error of" $g$

Rightarrow = 2% + 4% = 6%

Then, let's consider $\sqrt{\frac{L}{g}}$.

This expression can be rewritten as ${\left(\frac{L}{g}\right)}^{\frac{1}{2}}$.

When raising values to a power, we multiply the percentage error by that power:

Rightarrow % "error of" $\sqrt{\frac{L}{g}}$ = frac(1)(2) times % "error of" $\frac{L}{g}$

Rightarrow % "error of" $\sqrt{\frac{L}{g}}$ = frac(1)(2) times 6%

Rightarrow % "error of" $\sqrt{\frac{L}{g}}$ = 3%

Now, let's consider $2 \pi \sqrt{\frac{L}{g}}$.

When a value is multiplied by a constant, we multiply the absolute uncertainty by the constant as well.

However, when dealing with percentage errors, we leave it as it is.

So we do not need to multiply the percentage error by the constant.

Rightarrow % "error of" $2 \pi \sqrt{\frac{L}{g}}$ = % "error of" $\sqrt{\frac{L}{g}}$

therefore % "error of" $2 \pi \sqrt{\frac{L}{g}}$ =3%

Therefore, the percentage error in the measurement of the time period of the pendulum is 3%#.