Question 50: Oscillations

Question

Using a simple pendulum experiment $g$ is determind by measuring its time period $T$. Which of the following plots represent the correct relation between the pendulum length $L$ and time period $T$ ?

Accepted Answer

The time period $(T)$ of a simple pendulum of length $L$ is given by the standard formula :
$$ \mathrm{T}=2 \pi \sqrt{\mathrm{~L} / \mathrm{g}} $$
Where $g$ is the acceleration due to gravity.
The graphs in the options plot $\frac{1}{\mathrm{~T}^2}$ versus L .
On rearranging the formula for the time period,
$$ \mathrm{T}^2=\frac{4 \pi^2 \mathrm{~L}}{\mathrm{~g}} $$
$$\Rightarrow $$ $\frac{1}{\mathrm{~T}^2}=\frac{\mathrm{g}}{4 \pi^2 \mathrm{~L}}$
$$\Rightarrow $$$$ \frac{1}{\mathrm{~T}^2}=\left(\frac{\mathrm{g}}{4 \pi^2}\right) \frac{1}{\mathrm{~L}} $$
Since $g$ and $4 \pi^2$ are constants, the term $\frac{g}{4 \pi^2}$ is just a constant value. Let $\frac{g}{4 \pi^2}=k$, where $k$ is any positive constant.
Therefore, the relationship is :
$$ \mathrm{y}=\mathrm{kx} $$
where $\mathrm{y}=\frac{1}{\mathrm{~T}^2}$ and $\mathrm{x}=\mathrm{L}$.
The mathematical equation $\mathrm{y}=\mathrm{kx}$ (or $\mathrm{xy}=$ constant) represents a rectangular hyperbola, where $\frac{1}{\mathrm{~T}^2} \rightarrow 0$ if $\mathrm{L} \rightarrow \infty$, i.e., x -axis is horizontal asymptote. Also, $\frac{1}{\mathrm{~T}^2} \rightarrow \infty$ if $\mathrm{L} \rightarrow 0$, i.e., y-axis is horizontal asymptote
So, the resulting graph is,
Therefore, the correct option is (A).

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Oscillations

Using a simple pendulum experiment g is determind by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T ?

Choose the correct answer: