If the length of a simple pendulum is decreased by 2%, find the percentage decrease in its period T, where `T = 2pi sqrt((l)/(g))`

Explanation

To solve the problem of finding the percentage decrease in the period \( T \) of a simple pendulum when its length \( l \) is decreased by 2%, we start with the formula for the period of a simple pendulum:

\[ T = 2\pi \sqrt{\frac{l}{g}} \]

where \( g \) is the acceleration due to gravity, which is a constant.

### Step 1: Differentiate the Period with Respect to Length

We will differentiate \( T \) with respect to \( l \). First, we can express \( T \) as:

\[ T = 2\pi \left( l^{1/2} \cdot g^{-1/2} \right) \]

Now, differentiating both sides with respect to \( l \):

\[ \frac{dT}{dl} = 2\pi \cdot \frac{1}{2} l^{-1/2} \cdot g^{-1/2} = \frac{\pi}{\sqrt{l \cdot g}} \]

### Step 2: Relate the Change in Period to the Change in Length

Using the relationship between the changes in \( T \) and \( l \):

\[ \frac{dT}{T} = \frac{1}{2} \frac{dl}{l} \]

### Step 3: Substitute the Percentage Change in Length

Given that the length \( l \) is decreased by 2%, we can express this as:

\[ \frac{dl}{l} = -0.02 \]

### Step 4: Substitute into the Equation

Substituting \( \frac{dl}{l} \) into the equation:

\[ \frac{dT}{T} = \frac{1}{2} \left(-0.02\right) = -0.01 \]

### Step 5: Calculate the Percentage Change in Period

To find the percentage change in the period \( T \):

\[ \frac{dT}{T} \times 100 = -0.01 \times 100 = -1\% \]

Thus, the percentage decrease in the period \( T \) is:

\[ \text{Percentage decrease in } T = 1\% \]

### Final Answer

The percentage decrease in the period \( T \) of the pendulum when its length is decreased by 2% is **1%**.

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