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

Solution

Physics sir · 4 steps

Square the time period formula and you get T² = 4π²/gL, a straight line through the origin.

1. Start with T. For a simple pendulum, T = 2π√(L/g). T goes as √L, so a plot of T against L would curve.

   T² = (4π²/g) L

2. Read the shape. At one place g is fixed, so 4π²/g is a constant. That makes T² ∝ L: the graph has the form y = mx, a straight line with positive slope. When L = 0, T² = 0, so the line passes through the origin.
3. Check the options. (a) and (b) fall as L grows. (c) rises but bends over and flattens. Only (d) is a straight line rising from the origin.
4. So. The correct graph is (d).

The idea to keep. When a graph question asks for T² against L, square the formula first. Any relation of the form y = kx is a straight line through the origin, and its slope here is 4π²/g.

Answer from the NTA final answer key. Solution by PrepWiser.