Question 51: Oscillations

Question

Choose the correct length (L) versus square of the time period ($\mathrm{T}^{2}$) graph for a simple pendulum executing simple harmonic motion.

Accepted Answer

The relationship between the length of a pendulum and its period is given by the equation:
$T = 2\pi \sqrt {{L \over g}} $
$ \Rightarrow $ $\mathrm{T}^{2} = \frac{4\pi^2}{g} \cdot \mathrm{L}$
where T is the period of the pendulum, g is the acceleration due to gravity, and L is the length of the pendulum. This equation shows that the square of the period is directly proportional to the length of the pendulum.
Therefore, a graph of T
2
vs. L will be a straight line with a positive slope, passing through the origin. The slope of the line will be equal to $\frac{4\pi^2}{g}$, and its intercept will be zero.
Note :
Graph of T vs. L will be a parabola not T
2
vs. L as in equation of parabola y
2
= 4ax, graph of y vs. x is parabola not y
2
vs. x. As y
2
vs. x is a straight line.

Answer

Answer

Answer

Answer

Oscillations

Choose the correct length (L) versus square of the time period (\mathrm{T}^{2}) graph for a simple pendulum executing simple harmonic motion.

Choose the correct answer: