The Question is from neet-physics oscillations-waves , and it was created by the author SUBHANI

Question

A simple pendulum of length $L$ has a time period $T$.  If the length is increased by $\Delta L$ such that $\Delta L << L$, the change in time period $\Delta T$ is approximately:

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [A]
## CONCEPT FRAMEWORK
	**Primary Concept**: Simple Pendulum Time Period
	**Related Topics**: Small Angle Approximation, Differential Calculus
	**Prerequisites**: Formula for time period of a simple pendulum, understanding of differentials.
## SOLUTION PATHWAY
**Given Information**:
	* Time period of a simple pendulum: $T = 2\pi\sqrt{\frac{L}{g}}$
	* Length is increased by $\Delta L$, such that $\Delta L << L$
	* Time period changes by $\Delta T$
**Method & Steps**:
	1.	Find the time period with changed length.
		Reasoning: The new time period will be slightly different due to the change in length.
		Formula/Rule: $T' = 2\pi\sqrt{\frac{L+\Delta L}{g}}$
	2.	Find the change in time period, $\Delta T = T' - T$
		Working: $\Delta T = 2\pi\sqrt{\frac{L+\Delta L}{g}} - 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{L}{g}}\left(\sqrt{1+\frac{\Delta L}{L}} - 1ight)$. Since $\Delta L << L$, we can use the binomial approximation $\sqrt{1+x} \approx 1 + \frac{x}{2}$ for $x << 1$. Thus, $\sqrt{1+\frac{\Delta L}{L}} \approx 1 + \frac{\Delta L}{2L}$. So, $\Delta T = 2\pi\sqrt{\frac{L}{g}} \left(1 + \frac{\Delta L}{2L} - 1ight) = 2\pi\sqrt{\frac{L}{g}} \frac{\Delta L}{2L}$
		Key Point: The binomial approximation is crucial because $\Delta L$ is much smaller than $L$.
	3.	Simplify using the original time period $T$.
		Result: $\Delta T = T \frac{\Delta L}{2L}$
		Verification: The units are consistent with a change in time period. The result matches the correct option.
 ## OPTION ANALYSIS
[A] ✓: This is the correct option derived in the solution pathway. The change in time period is indeed $\frac{T\Delta L}{2L}$
[B] :x: This option has a factor of 2 in the numerator, which is incorrect.
[C] :x:  This option is missing the factor of 1/2.  It would be correct if the change in length was linear with respect to the change in time period, which is not the case due to the square root in the time period formula.
[D] :x: This is incorrect, and is missing the factor of the original time period $T$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting the binomial approximation. This will lead to an incorrect expression for the change in time period.
		2.[Error 2]: Not recognizing the original time period T in the final expression. This can lead to an answer with the correct relationship but incorrect form.
	**Quick Tips**:
		1.[Helpful shortcut]:  When dealing with small changes, differentials or binomial expansions are often helpful.
		2.[Memory aid]: Remember the time period formula $T=2\pi\sqrt{L/g}$ and the binomial approximation $(1+x)^n \approx 1+nx$ for small x.

SUBHANI · Answer

The correct Option is "\frac{T\Delta L}{2L}"

SUBHANI · Answer

\frac{2T\Delta L}{L}

SUBHANI · Answer

\frac{T\Delta L}{L}

SUBHANI · Answer

\frac{\Delta L}{2L}

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