A string of length 2 m is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of 500 Hz, then the waves would travel on its with a velocity of

Options

A. 125 `ms^(-1)`
B. 250 `ms^(-1)`
C. 500 `ms^(-1)`
D. 1000 `ms^(-1)`

Answer

To find the velocity of the wave on a string fixed at both ends vibrating in its fourth normal mode, we can follow these steps:

### Step 1: Understand the relationship between frequency, wave velocity, and string length.
The formula relating frequency (f), wave velocity (v), and the length of the string (L) for a string fixed at both ends is given by:

\[ f = \frac{n \cdot v}{2L} \]

where:
- \( f \) is the frequency of the wave,
- \( n \) is the mode number (1 for the first harmonic, 2 for the second harmonic, etc.),
- \( v \) is the wave velocity,
- \( L \) is the length of the string.

### Step 2: Identify the known values.
From the problem statement:
- Length of the string, \( L = 2 \, \text{m} \)
- Frequency, \( f = 500 \, \text{Hz} \)
- Mode number, \( n = 4 \) (since it is the fourth normal mode)

### Step 3: Rearrange the formula to solve for wave velocity (v).
We can rearrange the formula to express wave velocity \( v \):

\[ v = \frac{2L \cdot f}{n} \]

### Step 4: Substitute the known values into the equation.
Now, substituting the known values into the rearranged formula:

\[ v = \frac{2 \cdot 2 \, \text{m} \cdot 500 \, \text{Hz}}{4} \]

### Step 5: Calculate the wave velocity.
Calculating the above expression:

\[ v = \frac{4 \cdot 500}{4} = 500 \, \text{m/s} \]

### Conclusion:
The velocity of the waves traveling on the string is \( 500 \, \text{m/s} \).

---