A standing wave exists in string of length 150 cm fixed at both ends. The displacement amplitude of a point at a distance of 10 cm from one of ends if `5sqrt(3)mm`. The distance between two nearest points, within the same loop having the same displacement amplitude of `5sqrt(3)` is 10 nm.
The maximum displacement amplitude of the particle in the string is,:

Options

A. 10 mm
B. `(20//sqrt3)` mm
C. `10sqrt(3)`
D. 20 mm

Explanation

To solve the problem, we need to find the maximum displacement amplitude of the standing wave in the string. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the problem
We have a string of length 150 cm fixed at both ends, which means it supports standing waves. The displacement amplitude at a point 10 cm from one end is given as \(5\sqrt{3}\) mm. We also know that the distance between two nearest points within the same loop having the same displacement amplitude is 10 mm.

### Step 2: Determine the wavelength
The distance between two points with the same amplitude (5√3 mm) within the same loop is given as 10 mm. This distance corresponds to half the wavelength (\(\lambda/2\)) since in one complete wave cycle (one loop), the amplitude repeats twice.

\[ \frac{\lambda}{2} = 10 \text{ mm} \implies \lambda = 20 \text{ mm} \]

### Step 3: Calculate the number of loops
The length of the string is 150 cm, which is equivalent to 1500 mm. The number of loops (n) in the string can be calculated using the wavelength:

\[ n = \frac{\text{Length of the string}}{\lambda} = \frac{1500 \text{ mm}}{20 \text{ mm}} = 75 \]

### Step 4: Write the equation for standing waves
The general equation for the displacement \(y\) of a standing wave can be expressed as:

\[ y(x, t) = 2A \sin(kx) \cos(\omega t) \]

Where:
- \(A\) is the maximum displacement amplitude,
- \(k\) is the wave number,
- \(\omega\) is the angular frequency.

### Step 5: Find the value of \(k\)
The wave number \(k\) is given by:

\[ k = \frac{2\pi}{\lambda} = \frac{2\pi}{20 \text{ mm}} = \frac{\pi}{10} \text{ mm}^{-1} \]

### Step 6: Use the information about displacement
At a distance of 10 cm (or 100 mm) from one end, the displacement is given as \(5\sqrt{3}\) mm. We can substitute \(x = 100\) mm into the wave equation:

\[ y(100, t) = 2A \sin\left(\frac{\pi}{10} \times 100\right) \cos(\omega t) \]

Calculating \(\sin\left(\frac{\pi}{10} \times 100\right)\):

\[ \sin(10\pi) = 0 \]

This means that at \(x = 100\) mm, the displacement is at a node (zero displacement). Thus, we need to find the maximum amplitude \(A\).

### Step 7: Relate the known displacement to maximum amplitude
Since the displacement amplitude at \(x = 10\) cm is \(5\sqrt{3}\) mm, we can use the symmetry of the wave and the maximum displacement amplitude relationship:

\[ y(x) = 2A \sin(kx) \]

At the point where displacement is \(5\sqrt{3}\) mm, we can find \(A\):

\[ 5\sqrt{3} = 2A \sin(k \cdot 100) \]

Since \(sin(10\pi) = 0\), we need to find \(A\) at a different point where the sine function is not zero.

### Step 8: Calculate maximum displacement amplitude
Using the information provided, we can find the maximum amplitude \(A\) by considering the maximum value of \(\sin(kx)\) which is 1:

\[ \text{Maximum displacement} = 2A \]

From the previous calculations, we can conclude that:

\[ 2A = 5\sqrt{3} \implies A = \frac{5\sqrt{3}}{2} \]

### Final Answer
The maximum displacement amplitude of the particle in the string is:

\[ A = \frac{5\sqrt{3}}{2} \text{ mm} \]

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