Dimensional Analysis of a Pendulum Equation

Question: The time period \(T\) of a simple pendulum is given by the modified equation: \[ T = k\,L^a\,g^b, \] where \(L\) is the pendulum's length, \(g\) is the gravitational acceleration, and \(k\) is a dimensionless constant.

Select the correct answer from the options below:

• (1) \(a=\frac{1}{2},\; b=-\frac{1}{2}\)
• (2) \(a=1,\; b=-\frac{1}{2}\)
• (3) \(a=\frac{1}{2},\; b=-1\)
• (4) \(a=1,\; b=-1\)

Select the Correct Option

(1) \(a=\frac{1}{2},\; b=-\frac{1}{2}\)

(2) \(a=1,\; b=-\frac{1}{2}\)

(3) \(a=\frac{1}{2},\; b=-1\)

(4) \(a=1,\; b=-1\)

Step‑by‑Step Explanation

1. Step 1: Write Down the Standard Relationship
   We know that the standard formula for the time period of a simple pendulum is: \[ T \propto \sqrt{\frac{L}{g}}. \] This implies that the dimensions of \(T\) are given by: \[ T \sim L^{\frac{1}{2}}\,g^{-\frac{1}{2}}. \]
2. Step 2: Compare with the Modified Equation
   The modified equation is: \[ T = k\,L^a\,g^b, \] where \(k\) is dimensionless. For dimensional consistency, we require: \[ a = \frac{1}{2} \quad \text{and} \quad b = -\frac{1}{2}. \]
3. Step 3: Conclusion
   Therefore, the correct values of the exponents are: \[ a=\frac{1}{2} \quad \text{and} \quad b=-\frac{1}{2}. \]
4. Final Answer: The correct option is (1).

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