Question

Choose the correct length \(( L )\) versus square of time period (T²) graph for a simple pendulum executing simple harmonic motion

• A.
• B.
• C.
• D.

Hint:

For graph-based questions, establish the mathematical relationship between the variables involved. This will help identify the correct graph representing that relationship.

Answer 1

The Correct Option is C

A simple pendulum executing simple harmonic motion has its time period \(T\) given by the formula:

\(T = 2\pi \sqrt{\frac{L}{g}}\)

where:

• \(L\) is the length of the pendulum, and
• \(g\) is the acceleration due to gravity.

Squaring both sides, we get:

\(T^2 = 4\pi^2 \frac{L}{g}\)

This can be rearranged to:

\(L = \frac{g}{4\pi^2}T^2\)

This equation represents a linear relationship between \(L\) and \(T^2\) with the slope being \(\frac{g}{4\pi^2}\).

Therefore, the correct graph should be a straight line with \(T^2\) on the x-axis and \(L\) on the y-axis.

The correct length \((L)\) versus square of time period \((T^2)\) graph is:

This graph accurately represents the linear relationship between \(L\) and \(T^2\) as derived from the formula for the period of a simple pendulum.