A 2.0m long string with a linear mass density of `5.2xx10^(-3) kg m^(-1)` and tension 52N has both of its ends fixed. It vibrates in a standing wave patten with four antinodes. Frequency of the vibraion is :-

Options

A. 75 Hz
B. 150 Hz
C. 100 Hz
D. 50 Hz

Answer

To find the frequency of the vibration of the string, we can follow these steps:

### Step 1: Understand the Problem
We have a string of length \( L = 2.0 \, \text{m} \) with both ends fixed. It has a linear mass density \( \mu = 5.2 \times 10^{-3} \, \text{kg/m} \) and is under a tension \( T = 52 \, \text{N} \). The string vibrates in a standing wave pattern with four antinodes.

### Step 2: Determine the Wavelength
In a standing wave with both ends fixed, the number of antinodes \( n \) is related to the wavelength \( \lambda \) by the formula:
\[ n = \frac{L}{\lambda/2} \]
Given that there are four antinodes, we can write:
\[ 4 = \frac{L}{\lambda/2} \]
This implies:
\[ L = 2 \times \lambda \quad \Rightarrow \quad \lambda = \frac{L}{2} \]
Substituting \( L = 2.0 \, \text{m} \):
\[ \lambda = \frac{2.0 \, \text{m}}{2} = 1.0 \, \text{m} \]

### Step 3: Calculate the Wave Velocity
The velocity \( v \) of a wave on a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values for tension \( T = 52 \, \text{N} \) and linear mass density \( \mu = 5.2 \times 10^{-3} \, \text{kg/m} \):
\[ v = \sqrt{\frac{52}{5.2 \times 10^{-3}}} \]
Calculating the denominator:
\[ 5.2 \times 10^{-3} = 0.0052 \, \text{kg/m} \]
Now, calculating \( v \):
\[ v = \sqrt{\frac{52}{0.0052}} = \sqrt{10000} = 100 \, \text{m/s} \]

### Step 4: Calculate the Frequency
The frequency \( f \) of the wave can be calculated using the relationship:
\[ f = \frac{v}{\lambda} \]
Substituting the values for \( v \) and \( \lambda \):
\[ f = \frac{100 \, \text{m/s}}{1.0 \, \text{m}} = 100 \, \text{Hz} \]

### Final Answer
Thus, the frequency of the vibration is:
\[ \boxed{100 \, \text{Hz}} \]

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