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Question

A spring-mass system oscillates with a period 'T'. If the spring is cut into 'n' identical parts and one of these parts is used with the same mass, the new period of oscillation will be:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Simple Harmonic Motion (SHM) and Period of Oscillation
	**Related Topics**: Spring Constant, Effective Spring Constant, Time Period
	**Prerequisites**: Understanding of SHM, Spring Constant, and Period calculation.
## SOLUTION PATHWAY
**Given Information**:
	* Initial period of spring-mass system: T
	* Spring is cut into n identical parts
	* One part is used with the same mass

**Method & Steps**:
	1. **Initial Period Calculation**:
		Reasoning: The period of oscillation for a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{m}{k}}$, where m is the mass and k is the spring constant.
		Formula/Rule: $T = 2\pi \sqrt{\frac{m}{k}}$
	2. **Spring Constant of the Cut Spring**:
		Working: When a spring is cut into 'n' identical parts, the spring constant of each part becomes 'nk'. This is because the stiffness of the spring is inversely proportional to its length.
		Key Point: The effective spring constant increases when the spring length is reduced.
	3. **New Period Calculation**:
		Working:  Let the new period be $T'$. Using the new spring constant (nk) in the period formula, we get $T' = 2\pi \sqrt{\frac{m}{nk}} = 2\pi \sqrt{\frac{m}{k}} \frac{1}{\sqrt{n}}$
		Result:  Since $T = 2\pi \sqrt{\frac{m}{k}}$, then $T' = \frac{T}{\sqrt{n}}$
		Verification: The new period is smaller than the initial period, indicating higher oscillation frequency as expected.
 ## OPTION ANALYSIS
[A] :x: Incorrect. It suggests the period increases by n times, which is opposite of the correct behavior.
[B] :x: Incorrect. It suggests the period decreases by n times, not by square root of n.
[C] ✓: Correct. The period decreases by a factor of the square root of n.
[D] :x: Incorrect. It suggests the period increases by square root of n times, which is incorrect.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Confusing the relationship between spring length and spring constant. Remember that shorter springs are stiffer and have larger spring constants.
		2.[Error 2]: Forgetting the square root in the period formula when calculating the change due to changing spring constant.
	**Quick Tips**:
		1.[Helpful shortcut]: When a spring is divided into 'n' equal parts, the spring constant of each part increases by a factor of 'n'.
		2.[Memory aid]: Shorter springs have shorter periods of oscillation for the same mass.

P.NARASIMHA · Answer

$nT$

P.NARASIMHA · Answer

$T/n$

P.NARASIMHA · Answer

The correct Option is "$T/\sqrt{n}$"

P.NARASIMHA · Answer

$\sqrt{n}T$

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