A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, `y(x,t)=(0.01 m) sin[(62.8 m^(-1)x] cos[(628 s^(-1))t]`. Assuming `p = 3.14`, the correct statement(s) is (are)

Options

A. The number of nodes is 5.
B. the length of the string is 0.25 m
C. The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m
D. The fundamental frequency is 100 Hz

Explanation

To solve the problem, we will analyze the given wave equation and extract the necessary information step by step.

### Step 1: Identify the wave equation
The wave equation given is:
\[ y(x,t) = (0.01 \, \text{m}) \sin(62.8 \, \text{m}^{-1} x) \cos(628 \, \text{s}^{-1} t) \]

### Step 2: Determine the wave number \( k \) and angular frequency \( \omega \)
From the equation, we can identify:
- The wave number \( k = 62.8 \, \text{m}^{-1} \)
- The angular frequency \( \omega = 628 \, \text{s}^{-1} \)

### Step 3: Calculate the wavelength \( \lambda \)
The relationship between wave number \( k \) and wavelength \( \lambda \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} \]
Substituting \( k = 62.8 \):
\[ \lambda = \frac{2 \times 3.14}{62.8} = \frac{6.28}{62.8} \approx 0.1 \, \text{m} \]

### Step 4: Calculate the length of the string \( L \)
In the fifth harmonic, the length of the string \( L \) is related to the wavelength by:
\[ L = \frac{5\lambda}{2} \]
Substituting the value of \( \lambda \):
\[ L = \frac{5 \times 0.1}{2} = 0.25 \, \text{m} \]

### Step 5: Determine the number of nodes
In a string vibrating in the fifth harmonic, the number of nodes \( N \) can be calculated as:
\[ N = 5 + 1 = 6 \]
This includes the two fixed ends.

### Step 6: Determine the maximum displacement (amplitude)
The maximum displacement (amplitude) from the equation is given as:
\[ A = 0.01 \, \text{m} \]

### Step 7: Calculate the fundamental frequency \( f_0 \)
The speed of the wave \( v \) can be calculated using:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{628}{62.8} = 10 \, \text{m/s} \]
The fundamental frequency \( f_0 \) is given by:
\[ f_0 = \frac{v}{2L} \]
Substituting \( L = 0.25 \):
\[ f_0 = \frac{10}{2 \times 0.25} = \frac{10}{0.5} = 20 \, \text{Hz} \]

### Summary of Results
1. Length of the string \( L = 0.25 \, \text{m} \)
2. Number of nodes \( N = 6 \)
3. Maximum displacement (amplitude) \( A = 0.01 \, \text{m} \)
4. Fundamental frequency \( f_0 = 20 \, \text{Hz} \)

### Correct Statements
- The length of the string is 0.25 m (True).
- The number of nodes is 6 (True).
- The maximum displacement at the midpoint is 0.01 m (True).
- The fundamental frequency is 20 Hz (False, if given as 100 Hz).