A strain of sound waves is propagated along an organ pipe and gets reflected from an open end . If the displacement amplitude of the waves (incident and reflected) are `0.002 cm` , the frequency is `1000 Hz` and wavelength is `40 cm`. Then , the displacement amplitude of vibration at a point at distance `10 cm` from the open end , inside the pipe is

Answer

To find the displacement amplitude of vibration at a point 10 cm from the open end of an organ pipe, we can follow these steps:

### Step 1: Understand the Problem
We have an organ pipe with sound waves propagating through it. The displacement amplitude of the incident and reflected waves is given as \(0.002 \, \text{cm}\), the frequency is \(1000 \, \text{Hz}\), and the wavelength is \(40 \, \text{cm}\). We need to find the displacement amplitude at a point \(10 \, \text{cm}\) from the open end of the pipe.

### Step 2: Convert Units
Convert the displacement amplitude from centimeters to meters for consistency in calculations:
\[ 0.002 \, \text{cm} = 0.00002 \, \text{m} \]

### Step 3: Write the Equation for Stationary Waves
The equation for stationary waves in an open organ pipe is given by:
\[ y(x, t) = 2a \cos\left(\frac{2\pi x}{\lambda}\right) \sin(2\pi ft) \]
Where:
- \(y\) is the displacement at position \(x\) and time \(t\),
- \(a\) is the amplitude of the incident wave,
- \(\lambda\) is the wavelength,
- \(f\) is the frequency.

### Step 4: Substitute Known Values
Here, the amplitude \(a\) is \(0.00002 \, \text{m}\), the wavelength \(\lambda\) is \(40 \, \text{cm} = 0.4 \, \text{m}\), and we want to find the displacement at \(x = 10 \, \text{cm} = 0.1 \, \text{m}\).

Substituting these values into the equation:
\[ y(0.1, t) = 2(0.00002) \cos\left(\frac{2\pi (0.1)}{0.4}\right) \sin(2\pi (1000)t) \]

### Step 5: Calculate the Cosine Term
Calculate the cosine term:
\[ \frac{2\pi (0.1)}{0.4} = \frac{2\pi}{4} = \frac{\pi}{2} \]
Thus,
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

### Step 6: Final Calculation
Since the cosine term is zero, the displacement amplitude at \(x = 0.1 \, \text{m}\) is:
\[ y(0.1, t) = 2(0.00002)(0) \sin(2\pi (1000)t) = 0 \]

### Conclusion
The displacement amplitude of vibration at a point \(10 \, \text{cm}\) from the open end inside the pipe is:
\[ \boxed{0 \, \text{m}} \]

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