A wave moves with a certain speed in a stretched string. The percentage change in tension required to increase the velocity by 1 %, is approximately

Options

A. 1% increase
B. 1% decrease
C. 2% increase
D. 2% decrease

Answer

To solve the problem of determining the percentage change in tension required to increase the velocity of a wave in a stretched string by 1%, we can follow these steps:

### Step 1: Understand the relationship between wave speed, tension, and linear mass density
The speed \( v \) of a wave on a stretched string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string. Since the problem states that the linear mass density \( \mu \) remains constant, we can focus on the relationship between speed and tension.

### Step 2: Rearranging the formula
From the formula, we can express tension in terms of wave speed:
\[ T = \mu v^2 \]

### Step 3: Taking logarithms
To analyze the percentage change, we can take logarithms of both sides:
\[ \log T = \log(\mu v^2) = \log \mu + 2 \log v \]

### Step 4: Differentiate both sides
Differentiating both sides with respect to \( v \):
\[ \frac{dT}{T} = 2 \frac{dv}{v} \]

### Step 5: Relate percentage changes
The left side \( \frac{dT}{T} \) represents the relative change in tension, and the right side \( \frac{dv}{v} \) represents the relative change in speed. If we denote the percentage change in speed as \( \Delta v \) (which is given as 1% or 0.01), we can express this as:
\[ \frac{dv}{v} = 0.01 \]

### Step 6: Substitute and solve for \( \frac{dT}{T} \)
Substituting \( \frac{dv}{v} \) into the differentiated equation:
\[ \frac{dT}{T} = 2 \times 0.01 = 0.02 \]

### Step 7: Convert to percentage
To convert this to a percentage change, we multiply by 100:
\[ \frac{dT}{T} \times 100 = 0.02 \times 100 = 2\% \]

### Conclusion
Thus, the percentage change in tension required to increase the velocity by 1% is approximately **2%**.

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