The period of a pendulum is given as T = 2 π (l/g)1/2 where g = 9.81 m/s2 and π = 3.1416. The period of a pendulum of length 1 m correct to the first place of decimal in seconds is
2.0
The question asks us to find the period of a simple pendulum with a given length, using a standard formula and specific values for gravity and pi. We need to calculate the period and round the result to the first decimal place.
The formula provided for the period ($T$) of a simple pendulum is:
$\text{T} = 2 \pi \sqrt{\frac{\text{l}}{\text{g}}}$
Where:
We are given the following values:
Now, we substitute the given values into the formula for the period:
$\text{T} = 2 \times 3.1416 \times \sqrt{\frac{1}{9.81}}$
First, calculate the value inside the square root:
$\frac{1}{9.81} \approx 0.1019368$
Next, calculate the square root of this value:
$\sqrt{0.1019368} \approx 0.319275$
Now, multiply the terms together:
$\text{T} = 2 \times 3.1416 \times 0.319275$
$\text{T} = 6.2832 \times 0.319275$
$\text{T} \approx 2.00581$ seconds
The question asks for the period correct to the first place of decimal. Our calculated value is approximately $2.00581$.
To round to the first decimal place, we look at the second decimal digit. If it is 5 or greater, we round up the first decimal digit. If it is less than 5, we keep the first decimal digit as it is.
In $2.00581$, the first decimal digit is 0, and the second decimal digit is 0. Since 0 is less than 5, we keep the first decimal digit as 0.
So, $2.00581$ rounded to the first decimal place is $2.0$.
The period of the pendulum of length 1 m, correct to the first place of decimal, is 2.0 seconds.
| Parameter | Value |
|---|---|
| Pendulum Length (l) | 1 m |
| Gravity (g) | 9.81 m/s² |
| Pi ($\pi$) | 3.1416 |
| Calculated Period (T) | $\approx 2.00581$ s |
| Rounded Period (T) | 2.0 s |
The calculated period of the pendulum, rounded to one decimal place, matches one of the provided options.
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