The Question is from NEET guide , and it was created by the author VENKATESWARARAO

Question

If the linear mass density of a string is quadrupled, keeping the tension constant, the speed of a transverse wave on the string will:

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: B
## CONCEPT FRAMEWORK
 **Primary Concept**: Wave speed on a string
 **Related Topics**: Linear mass density, tension
 **Prerequisites**: Understanding of wave mechanics
## SOLUTION PATHWAY
**Given Information**:
  Linear mass density is quadrupled ($\mu_{new} = 4\mu_{old}$)
  Tension remains constant ($T_{new} = T_{old}$)
**Method & Steps**:
 1. The speed of a transverse wave on a string is given by the formula:
  Reasoning: This formula relates the wave speed to the tension and linear mass density of the string.
  Formula/Rule: $v = \sqrt{\frac{T}{\mu}}$ where $v$ is the wave speed, $T$ is the tension, and $\mu$ is the linear mass density.
 2. Let's denote the initial speed as $v_{old}$ and the new speed as $v_{new}$. We have:
  Working: $v_{old} = \sqrt{\frac{T_{old}}{\mu_{old}}}$ and $v_{new} = \sqrt{\frac{T_{new}}{\mu_{new}}} = \sqrt{\frac{T_{old}}{4\mu_{old}}}$
  Key Point:  Since $T_{new} = T_{old}$, we can directly substitute the new linear mass density.
 3. Comparing the two speeds:
  Result: $v_{new} = \sqrt{\frac{T_{old}}{4\mu_{old}}} = \frac{1}{2}\sqrt{\frac{T_{old}}{\mu_{old}}} = \frac{1}{2}v_{old}$ 
  Verification: The new speed is half the original speed.
 ## OPTION ANALYSIS
A] :x:  Doubling the speed is incorrect. The speed is inversely proportional to the square root of the linear mass density. 
B] ✓: Halving the speed is correct, as shown in the calculation. 
C] :x: The speed does not remain the same; it changes with the linear mass density. 
D] :x: The speed does not become four times. It is inversely proportional to the square root of the linear mass density. 
 ## LEARNING AIDS
 **Common Mistakes**:
  1. Forgetting to take the square root:  Students might mistakenly think the speed is simply inversely proportional to the linear mass density, instead of its square root.
  2. Incorrectly relating tension and linear mass density:  Students might incorrectly assume a direct proportionality between speed and either tension or linear mass density.
 **Quick Tips**:
  1. Remember the formula $v = \sqrt{\frac{T}{\mu}}$ and focus on the relationship between each term and the speed. 
  2. Visualize the wave's motion: A heavier string (higher μ) will have slower wave propagation.

VENKATESWARARAO · Answer

Become double

VENKATESWARARAO · Answer

The correct Option is "Become half"

VENKATESWARARAO · Answer

Remain the same

VENKATESWARARAO · Answer

Become four times

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