Three consecutive resonant frequencies of a string are `90, 150` and `210 Hz`. If the length of the string is `80 cm`, what would be the seed (in `m//s`) of a transverse wave on this string ?

Options

A. 45m/s
B. 75m/s
C. 48m/s
D. 80m/s

Explanation

To find the speed of a transverse wave on a string given its resonant frequencies and length, we can follow these steps:

### Step 1: Identify the given frequencies and length
The three consecutive resonant frequencies are:
- \( f_1 = 90 \, \text{Hz} \)
- \( f_2 = 150 \, \text{Hz} \)
- \( f_3 = 210 \, \text{Hz} \)

The length of the string is:
- \( L = 80 \, \text{cm} = 0.8 \, \text{m} \)

### Step 2: Understand the relationship between frequencies and harmonics
For a string fixed at one end and free at the other, the frequencies of the harmonics can be expressed as:
- \( f_n = \frac{nV}{2L} \)
Where \( n \) is the harmonic number, \( V \) is the speed of the wave, and \( L \) is the length of the string.

### Step 3: Set up equations for the frequencies
For the first frequency \( f_1 \):
\[ f_1 = \frac{nV}{2L} = 90 \, \text{Hz} \]
For the second frequency \( f_2 \):
\[ f_2 = \frac{(n+2)V}{2L} = 150 \, \text{Hz} \]

### Step 4: Divide the equations to eliminate \( V \)
Dividing the first equation by the second:
\[ \frac{f_1}{f_2} = \frac{n}{n+2} \]
Substituting the values:
\[ \frac{90}{150} = \frac{n}{n+2} \]
This simplifies to:
\[ \frac{3}{5} = \frac{n}{n+2} \]

### Step 5: Cross-multiply and solve for \( n \)
Cross-multiplying gives:
\[ 3(n + 2) = 5n \]
Expanding and rearranging:
\[ 3n + 6 = 5n \implies 2n = 6 \implies n = 3 \]

### Step 6: Substitute \( n \) back to find \( V \)
Now substitute \( n = 3 \) back into the first frequency equation:
\[ 90 = \frac{3V}{2 \times 0.8} \]
Solving for \( V \):
\[ 90 = \frac{3V}{1.6} \implies 90 \times 1.6 = 3V \implies 144 = 3V \implies V = \frac{144}{3} = 48 \, \text{m/s} \]

### Final Answer
The speed of the transverse wave on the string is:
\[ \boxed{48 \, \text{m/s}} \]