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Question

A transverse wave travels on a string with a speed of 200 m/s. If the string's linear mass density is doubled while the tension remains constant, what will be the new speed of the wave?

SUBHANI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
 **Primary Concept**: Wave speed on a string
 **Related Topics**: Linear mass density, tension
 **Prerequisites**: Understanding of wave properties and the relationship between wave speed, tension, and linear mass density.
## SOLUTION PATHWAY
**Given Information**:
  Initial wave speed, $v_1 = 200 m/s$
  Initial linear mass density, $μ_1 = μ$
  Final linear mass density, $μ_2 = 2μ$
  Tension remains constant, $T_1 = T_2 = T$
**Method & Steps**:
 1. The speed of a transverse wave on a string is given by the formula:
  $v = \sqrt{\frac{T}{μ}}$
  where $v$ is the wave speed, $T$ is the tension in the string, and $μ$ is the linear mass density.
  Reasoning: This formula shows that the wave speed is inversely proportional to the square root of the linear mass density when the tension is constant.
  Formula/Rule: $v = \sqrt{\frac{T}{μ}}$
 2. Let's find the ratio of the final speed ($v_2$) to the initial speed ($v_1$):
  $\frac{v_2}{v_1} = \frac{\sqrt{\frac{T}{μ_2}}}{\sqrt{\frac{T}{μ_1}}} = \sqrt{\frac{μ_1}{μ_2}} = \sqrt{\frac{μ}{2μ}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
  Working: Substituting the given values into the formula.
  Key Point:  The tension remains constant, so it cancels out in the ratio.
 3. Now we can calculate the new speed:
  $v_2 = v_1 	imes \frac{1}{\sqrt{2}} = 200 m/s 	imes \frac{1}{\sqrt{2}} \approx 141 m/s$
  Result: The new speed is approximately 141 m/s.
  Verification: The new speed should be less than the initial speed since the linear mass density has increased.
 ## OPTION ANALYSIS
A :x: 100 m/s (Incorrect calculation)
B ✓: 141 m/s (Correct calculation)
C :x: 200 m/s (Ignoring the change in linear mass density)
D :x: 282 m/s (Incorrect calculation)
 ## LEARNING AIDS
 **Common Mistakes**:
  1.Incorrectly using the formula for wave speed, perhaps forgetting the square root or inverting the relationship between speed and linear mass density.: To avoid this, carefully write out the formula and substitute values correctly.
  2.Forgetting that the speed is inversely proportional to the square root of linear mass density: Double-check the relationship between the variables in the formula.
 **Quick Tips**:
  1.When the tension is constant, focus on the relationship between speed and linear mass density:  $v \propto \frac{1}{\sqrt{μ}}$
  2.Remember that doubling the linear mass density will reduce the wave speed by a factor of $\frac{1}{\sqrt{2}}$

SUBHANI · Answer

100 m/s

SUBHANI · Answer

The correct Option is "141 m/s"

SUBHANI · Answer

200 m/s

SUBHANI · Answer

282 m/s

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