Oscillations

JEE Main level, single correct, medium

149 of 194

Question 149 of 194

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

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Answer: (B)

The time period TT of a simple pendulum of length LL is given by the formula:
T=2πLgT = 2\pi \sqrt{\frac{L}{g}}
Squaring both sides, we get:

T2=4π2LgT^2 = 4\pi^2 \frac{L}{g}
Rearranging the equation to find the relationship between
1T2\frac{1}{T^2} and LL:
1T2=g4π2L\frac{1}{T^2} = \frac{g}{4\pi^2 L}
Let
y=1T2y = \frac{1}{T^2} and x=Lx = L. The equation becomes:
y=kxy = \frac{k}{x} where k=g4π2k = \frac{g}{4\pi^2} is a constant.
This is the equation of a rectangular hyperbola in the first quadrant.

As
LL increases, 1T2\frac{1}{T^2} decreases non-linearly, approaching the axes asymptotically.
Comparing this with the given plots, option (2) correctly represents a hyperbolic relationship between
1T2\frac{1}{T^2} and LL.

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