A one-metre long stretched string having a mass of 40 g is attached to a tuning fork. The fork vibrates at 128 Hz in a direction perpendicular to the string. What should be the tension in the string if it is to vibrate in four loops ?

Explanation

To find the tension in the string when it vibrates in four loops, we can follow these steps:

### Step 1: Understand the relationship between loops and wavelength
In a string vibrating in four loops, the length of the string (L) is equal to 2 wavelengths (λ). Therefore, we can express this relationship as:
\[ L = 2\lambda \]
Given that the length of the string is 1 meter, we can write:
\[ 1 = 2\lambda \]
From this, we can solve for the wavelength:
\[ \lambda = \frac{1}{2} \text{ meter} = 0.5 \text{ meter} \]

### Step 2: Calculate the wave velocity
The wave velocity (v) on the string can be calculated using the frequency (f) and the wavelength (λ):
\[ v = f \cdot \lambda \]
Given that the frequency of the tuning fork is 128 Hz:
\[ v = 128 \cdot 0.5 = 64 \text{ m/s} \]

### Step 3: Determine the mass per unit length (μ)
The mass of the string is given as 40 g, which we need to convert to kilograms:
\[ \text{mass} = 40 \text{ g} = 0.04 \text{ kg} \]
The mass per unit length (μ) is then:
\[ \mu = \frac{\text{mass}}{\text{length}} = \frac{0.04 \text{ kg}}{1 \text{ m}} = 0.04 \text{ kg/m} \]

### Step 4: Relate tension (T) to wave velocity and mass per unit length
The relationship between tension (T), wave velocity (v), and mass per unit length (μ) is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Squaring both sides gives:
\[ v^2 = \frac{T}{\mu} \]
From this, we can solve for tension (T):
\[ T = v^2 \cdot \mu \]

### Step 5: Substitute the values to find tension
Now we can substitute the values we have found for v and μ:
\[ T = (64 \text{ m/s})^2 \cdot (0.04 \text{ kg/m}) \]
Calculating this gives:
\[ T = 4096 \cdot 0.04 = 163.84 \text{ N} \]

### Final Answer
Thus, the tension in the string should be approximately:
\[ T \approx 163.84 \text{ N} \]

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