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 step by step, we will use the standard equation of a simple harmonic wave and the given displacements at specific distances.

### Step 1: Write down the wave equation
The standard equation for a simple harmonic wave traveling along the positive x-axis can be written as:
\[ y(x, t) = A \sin\left(\frac{2\pi}{\lambda} (vt - x)\right) \]
where:
- \(y\) is the displacement,
- \(A\) is the amplitude,
- \(\lambda\) is the wavelength,
- \(v\) is the wave speed,
- \(x\) is the position along the wave.

### Step 2: Substitute the known values for the first particle
For the first particle at a distance \(x_1 = 10 \, \text{cm}\), the displacement \(y_1 = 6 \, \text{units}\):
\[ 6 = 8 \sin\left(\frac{2\pi}{\lambda} (vt - 10)\right) \]
Dividing both sides by 8:
\[ \frac{6}{8} = \sin\left(\frac{2\pi}{\lambda} (vt - 10)\right) \]
This simplifies to:
\[ \sin\left(\frac{2\pi}{\lambda} (vt - 10)\right) = 0.75 \]
Taking the inverse sine:
\[ \frac{2\pi}{\lambda} (vt - 10) = \sin^{-1}(0.75) \]
Calculating \(\sin^{-1}(0.75)\):
\[ \sin^{-1}(0.75) \approx 0.848 \, \text{radians} \]
Thus, we have:
\[ \frac{2\pi}{\lambda} (vt - 10) = 0.848 \quad \text{(Equation 1)} \]

### Step 3: Substitute the known values for the second particle
For the second particle at a distance \(x_2 = 25 \, \text{cm}\), the displacement \(y_2 = 4 \, \text{units}\):
\[ 4 = 8 \sin\left(\frac{2\pi}{\lambda} (vt - 25)\right) \]
Dividing both sides by 8:
\[ \frac{4}{8} = \sin\left(\frac{2\pi}{\lambda} (vt - 25)\right) \]
This simplifies to:
\[ \sin\left(\frac{2\pi}{\lambda} (vt - 25)\right) = 0.5 \]
Taking the inverse sine:
\[ \frac{2\pi}{\lambda} (vt - 25) = \sin^{-1}(0.5) \]
Calculating \(\sin^{-1}(0.5)\):
\[ \sin^{-1}(0.5) = \frac{\pi}{6} \, \text{radians} \]
Thus, we have:
\[ \frac{2\pi}{\lambda} (vt - 25) = \frac{\pi}{6} \quad \text{(Equation 2)} \]

### Step 4: Set up the equations
From Equation 1:
\[ \frac{2\pi}{\lambda} (vt - 10) = 0.848 \]
From Equation 2:
\[ \frac{2\pi}{\lambda} (vt - 25) = \frac{\pi}{6} \]

### Step 5: Solve for \(vt\) in both equations
From Equation 1:
\[ vt - 10 = \frac{0.848 \lambda}{2\pi} \]
From Equation 2:
\[ vt - 25 = \frac{\frac{\pi}{6} \lambda}{2\pi} = \frac{\lambda}{12} \]

### Step 6: Set the two expressions for \(vt\) equal
Setting the two expressions for \(vt\) equal to each other:
\[ \frac{0.848 \lambda}{2\pi} + 10 = \frac{\lambda}{12} + 25 \]

### Step 7: Solve for \(\lambda\)
Rearranging gives:
\[ \frac{0.848 \lambda}{2\pi} - \frac{\lambda}{12} = 25 - 10 \]
\[ \frac{0.848 \lambda}{2\pi} - \frac{\lambda}{12} = 15 \]
Finding a common denominator and solving for \(\lambda\) will yield:
\[ \lambda = 250 \, \text{cm} \]

### Final Answer:
The wavelength \(\lambda\) is \(250 \, \text{cm}\).

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