A string fixed at both ends, oscillate in 4th harmonic. The displacement of particular wave is given as `Y=2Asin(5piX)cos(100pit)`. Then find the length of the string?

Options

A. 80 cm
B. 100cm
C. 60 cm
D. 120 cm

Answer

To find the length of the string that is oscillating in its 4th harmonic, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Harmonic Nature**:
The string is fixed at both ends and is vibrating in its 4th harmonic. In the nth harmonic, the number of loops formed is equal to n. Therefore, for the 4th harmonic, there are 4 loops.

2. **Relate Length and Wavelength**:
The length of the string (L) can be related to the wavelength (λ) by the formula:
\[ L = n \cdot \frac{\lambda}{2} \]
where n is the harmonic number. For the 4th harmonic (n = 4), we have:
\[ L = 4 \cdot \frac{\lambda}{2} = 2\lambda \]

3. **Identify the Wave Equation**:
The displacement of the wave is given by:
\[ Y = 2A \sin(5\pi x) \cos(100\pi t) \]
This can be compared to the standard form of a standing wave:
\[ Y = 2A \sin(kx) \cos(\omega t) \]
From this, we can identify the wave number \( k \) and angular frequency \( \omega \).

4. **Extract the Wave Number**:
From the equation, we see that:
\[ k = 5\pi \]
The wave number \( k \) is related to the wavelength \( \lambda \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]

5. **Calculate the Wavelength**:
Setting the expressions for \( k \) equal gives:
\[ 5\pi = \frac{2\pi}{\lambda} \]
Solving for \( \lambda \):
\[ \lambda = \frac{2\pi}{5\pi} = \frac{2}{5} = 0.4 \text{ meters} = 40 \text{ cm} \]

6. **Calculate the Length of the String**:
Now substituting \( \lambda \) back into the length formula:
\[ L = 2\lambda = 2 \cdot 40 \text{ cm} = 80 \text{ cm} \]

### Final Answer:
The length of the string is **80 cm**.