Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length ($l$), mass of the bob ($m$) and acceleration due to gravity ($g$). Derive

Question

Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length ($l$), mass of the bob ($m$) and acceleration due to gravity ($g$). Derive the expression for its time period using method of dimensions.

Accepted Answer

> [!TLDR]
> Dimensional analysis reveals that the period of a simple pendulum depends only on its length and gravity, not mass. The result is $T = k\sqrt{\frac{l}{g}}$, where $k$ is a dimensionless constant (which turns out to be $2\pi$).

The method of dimensions rests on a powerful principle: any physically meaningful equation must be dimensionally consistent. When we don't know the exact form of a relationship but understand which variables matter, we can deduce the structure of the formula by demanding that both sides have the same dimensions.

For a simple pendulum, we suspect the period $T$ depends on length $l$, mass $m$, and acceleration due to gravity $g$. We assume a power-law relationship:

$$T = k \cdot l^a \cdot m^b \cdot g^c$$

where $k$ is a dimensionless constant and $a$, $b$, $c$ are exponents we need to find.

## Step-by-step derivation

**1. Write down the dimensions of each quantity**

Every physical quantity can be expressed in terms of fundamental dimensions: mass $[M]$, length $[L]$, and time $[T]$.

| Quantity | Symbol | Dimensions |
|----------|--------|------------|
| Period | $T$ | $[T]$ |
| Length | $l$ | $[L]$ |
| Mass | $m$ | $[M]$ |
| Acceleration | $g$ | $[LT^{-2}]$ |

**2. Substitute dimensions into the assumed relationship**

Replace each variable with its dimensional formula:

$$[T] = [L]^a \cdot [M]^b \cdot [LT^{-2}]^c$$

Simplify the right side:

$$[T] = [L]^a \cdot [M]^b \cdot [L]^c \cdot [T]^{-2c}$$

$$[T] = [M]^b \cdot [L]^{a+c} \cdot [T]^{-2c}$$

**3. Equate the exponents of corresponding dimensions**

For dimensional consistency, the exponents of $[M]$, $[L]$, and $[T]$ on both sides must match.

Left side: $[M]^0 \cdot [L]^0 \cdot [T]^1$

Right side: $[M]^b \cdot [L]^{a+c} \cdot [T]^{-2c}$

This gives us three equations:

- For $[M]$: $b = 0$
- For $[L]$: $a + c = 0$
- For $[T]$: $-2c = 1$

**4. Solve the system of equations**

From the time equation:
$$-2c = 1 \implies c = -\frac{1}{2}$$

From the length equation:
$$a + c

Concept understanding — Significant Figures Calculation

Significant Figures: The Art of Honest Measurement

The Core Idea

The Rules (Memorize These)

1. Non-zero digits are always significant

2. Zeros between non-zero digits are significant

3. Leading zeros are never significant

4. Trailing zeros are significant only if there's a decimal point

5. Exact numbers have infinite sig figs

Why This Matters: Calculations

Multiplication and Division

Addition and Subtraction

Why this formula?

Significant Figures: Why the Rules Work

1. The Fundamental Idea: Uncertainty is the Key

2. Rule for Addition and Subtraction

Why this holds

3. Rule for Multiplication and Division

Why this holds

4. Why These Rules Are Different …

Step-by-step derivation

Concept: Dimensional Analysis

Common Mistakes in Deriving the Time Period of a Simple Pendulum by Dimensional Analysis

Mistake 1: Assuming the period must depend on mass, since the bob has mass

Mistake 2: Mixing up the exponent equations for $L$ and $T$

Mistake 3: Getting the sign of $c$ wrong

Mistake 4: Thinking dimensional analysis gives the constant $k$

Quantity	Symbol	Dimensions
Period	$T$	$[T]$
Length	$l$	$[L]$
Mass	$m$	$[M]$
Acceleration	$g$	$[LT^{-2}]$

Step-by-step derivation

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