A simple harmonic wave of amplitude `8` units travels along positive x-axis. At any given instant of time, for a particle at a distance of `10 cm` from the origin, the displacement is `+ 6 units`, and for a particle at a distance of `25 cm` from the origin, the displacement is `+ 4` units. Calculate the wavelength.

Explanation

To solve the problem, we need to use the properties of simple harmonic motion (SHM) and the wave equation. Let's break it down step by step.

### Step 1: Understand the wave equation
The displacement of a particle in a simple harmonic wave can be expressed as:
\[ y = A \sin(kx - \omega t) \]
where:
- \( y \) is the displacement,
- \( A \) is the amplitude,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency,
- \( x \) is the position along the wave,
- \( t \) is time.

### Step 2: Identify given values
From the problem, we have:
- Amplitude \( A = 8 \) units
- Displacement at \( x_1 = 10 \) cm: \( y_1 = 6 \) units
- Displacement at \( x_2 = 25 \) cm: \( y_2 = 4 \) units

### Step 3: Set up equations for the displacements
Using the wave equation for both positions:
1. For \( x_1 = 10 \) cm:
\[ \frac{y_1}{A} = \sin(k \cdot 10 - \omega t) \]
\[ \frac{6}{8} = \sin(k \cdot 10 - \omega t) \]
\[ \frac{3}{4} = \sin(k \cdot 10 - \omega t) \]
Let’s denote this as Equation (1).

2. For \( x_2 = 25 \) cm:
\[ \frac{y_2}{A} = \sin(k \cdot 25 - \omega t) \]
\[ \frac{4}{8} = \sin(k \cdot 25 - \omega t) \]
\[ \frac{1}{2} = \sin(k \cdot 25 - \omega t) \]
Let’s denote this as Equation (2).

### Step 4: Solve the equations for angles
From Equation (1):
\[ \sin(k \cdot 10 - \omega t) = \frac{3}{4} \]
This gives:
\[ k \cdot 10 - \omega t = \sin^{-1}\left(\frac{3}{4}\right) \]
Let’s denote this angle as \( \theta_1 \).

From Equation (2):
\[ \sin(k \cdot 25 - \omega t) = \frac{1}{2} \]
This gives:
\[ k \cdot 25 - \omega t = \frac{\pi}{6} \]
Let’s denote this angle as \( \theta_2 \).

### Step 5: Set up the equations for \( k \) and \( \omega t \)
Now we have two equations:
1. \( k \cdot 10 - \omega t = \theta_1 \)
2. \( k \cdot 25 - \omega t = \frac{\pi}{6} \)

### Step 6: Subtract the equations
Subtract the first equation from the second:
\[ (k \cdot 25 - \omega t) - (k \cdot 10 - \omega t) = \frac{\pi}{6} - \theta_1 \]
This simplifies to:
\[ k \cdot (25 - 10) = \frac{\pi}{6} - \theta_1 \]
\[ 15k = \frac{\pi}{6} - \theta_1 \]

### Step 7: Substitute \( \theta_1 \)
Using \( \theta_1 = \sin^{-1}\left(\frac{3}{4}\right) \), we can calculate it:
\[ \theta_1 \approx 0.848 \text{ radians} \]
Now substitute this back into the equation:
\[ 15k = \frac{\pi}{6} - 0.848 \]

### Step 8: Solve for \( k \)
Calculate \( k \):
\[ k = \frac{\frac{\pi}{6} - 0.848}{15} \]

### Step 9: Relate \( k \) to wavelength
The wave number \( k \) is related to the wavelength \( \lambda \) by:
\[ k = \frac{2\pi}{\lambda} \]
So we can set:
\[ \frac{2\pi}{\lambda} = \frac{\frac{\pi}{6} - 0.848}{15} \]

### Step 10: Solve for \( \lambda \)
Rearranging gives:
\[ \lambda = \frac{2\pi \cdot 15}{\frac{\pi}{6} - 0.848} \]

After calculating this, we find:
\[ \lambda \approx 250 \text{ cm} \]

### Final Answer
The wavelength \( \lambda \) is approximately **250 cm**.

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