The speed of a wave on a string is 150 `"ms"^(-1)` when the tension is 120 N. The percentage increase in the tension in order to raise the wave speed by `20%` is

Options

A. `44%`
B. `40%`
C. `20%`
D. `10%`

Explanation

To solve the problem of finding the percentage increase in tension required to raise the wave speed by 20%, we can follow these steps:

### Step 1: Understand the relationship between wave speed and tension
The speed of a wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( v \) is the wave speed, \( T \) is the tension, and \( \mu \) is the mass per unit length of the string. For this problem, we will assume that \( \mu \) remains constant.

### Step 2: Calculate the new wave speed
The initial speed \( v_1 \) is given as 150 m/s. We need to find the new speed \( v_2 \) after a 20% increase:
\[ v_2 = v_1 + 0.2 \times v_1 = 150 + 0.2 \times 150 = 150 + 30 = 180 \, \text{m/s} \]

### Step 3: Set up the ratio of speeds and tensions
From the relationship between speed and tension, we can write:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T_1}{T_2}} \]
where \( T_1 \) is the initial tension (120 N) and \( T_2 \) is the new tension we want to find.

### Step 4: Substitute known values into the equation
Substituting the known values:
\[ \frac{150}{180} = \sqrt{\frac{120}{T_2}} \]

### Step 5: Simplify the ratio
Simplifying the left side:
\[ \frac{150}{180} = \frac{5}{6} \]
So we have:
\[ \frac{5}{6} = \sqrt{\frac{120}{T_2}} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \left(\frac{5}{6}\right)^2 = \frac{120}{T_2} \]
\[ \frac{25}{36} = \frac{120}{T_2} \]

### Step 7: Solve for \( T_2 \)
Cross-multiplying gives:
\[ 25 T_2 = 36 \times 120 \]
\[ T_2 = \frac{36 \times 120}{25} \]
Calculating this:
\[ T_2 = \frac{4320}{25} = 172.8 \, \text{N} \]

### Step 8: Calculate the percentage increase in tension
The percentage increase in tension is given by:
\[ \text{Percentage Increase} = \frac{T_2 - T_1}{T_1} \times 100 \]
Substituting the values:
\[ \text{Percentage Increase} = \frac{172.8 - 120}{120} \times 100 \]
\[ = \frac{52.8}{120} \times 100 \]
\[ = 44\% \]

### Final Answer
The percentage increase in tension required to raise the wave speed by 20% is **44%**.

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