The speed of a transverse wave on a string is 115 m/s when the string tension is 200 N. To what value must the tension be changed to raise the wave speed to 223 m/s?

Answer

To solve the problem, we need to use the relationship between the speed of a transverse wave on a string, the tension in the string, and the mass per unit length of the string. The formula for the speed of a transverse wave is given by:

\[ v = \sqrt{\frac{T}{\mu}} \]

where:
- \( v \) is the speed of the wave,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

Since the mass per unit length \( \mu \) remains constant (as we are not changing the string), we can establish a relationship between the initial and final speeds and tensions.

### Step 1: Write the relationship between the speeds and tensions

From the formula, we can express the relationship as:

\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]

### Step 2: Substitute the known values

Given:
- Initial speed \( v_1 = 115 \, \text{m/s} \)
- Final speed \( v_2 = 223 \, \text{m/s} \)
- Initial tension \( T_1 = 200 \, \text{N} \)

Substituting these values into the equation:

\[ \frac{115}{223} = \sqrt{\frac{200}{T_2}} \]

### Step 3: Square both sides to eliminate the square root

Squaring both sides gives:

\[ \left(\frac{115}{223}\right)^2 = \frac{200}{T_2} \]

### Step 4: Calculate the left side

Calculating \( \left(\frac{115}{223}\right)^2 \):

\[ \frac{115^2}{223^2} = \frac{13225}{49729} \approx 0.2659 \]

### Step 5: Set up the equation to solve for \( T_2 \)

Now we have:

\[ 0.2659 = \frac{200}{T_2} \]

### Step 6: Rearrange to find \( T_2 \)

Rearranging the equation gives:

\[ T_2 = \frac{200}{0.2659} \]

### Step 7: Calculate \( T_2 \)

Calculating \( T_2 \):

\[ T_2 \approx 752.19 \, \text{N} \]

### Step 8: Round the answer

Rounding to the nearest whole number, we get:

\[ T_2 \approx 752 \, \text{N} \]

### Final Answer

The tension must be changed to approximately **752 N** to raise the wave speed to 223 m/s.

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