Question

A sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz when kept under some tension. The resonating length of the wire with fundamental frequency of 600 Hz under same tension _______ cm.

Answer 1

Correct Answer: 60

Given:
- Initial resonating length, \( L = 90 \, \text{cm} \)
- Initial fundamental frequency, \( f_0 = 400 \, \text{Hz} \)
- New fundamental frequency, \( f' = 600 \, \text{Hz} \)

Step 1: Relation Between Frequency and Length
The fundamental frequency of a vibrating string is given by:

\[ f_0 = \frac{v}{2L}, \]

where:
- \( v \) is the wave speed,
- \( L \) is the length of the wire.
For the same tension, the wave speed \( v \) remains constant.

Step 2: Expressing New Length in Terms of Frequency
Let the new resonating length be \( L' \) for the frequency \( f' \). The new fundamental frequency is given by:

\[ f' = \frac{v}{2L'}. \]

Dividing the two equations:

\[ \frac{f'}{f_0} = \frac{L}{L'}. \]

Rearranging to find \( L' \):

\[ L' = L \times \frac{f_0}{f'}. \]

Step 3: Substituting the Given Values
Substituting the values:

\[ L' = 90 \times \frac{400}{600}. \]

Simplifying:

\[ L' = 90 \times \frac{2}{3} = 60 \, \text{cm}. \]

Therefore, the new resonating length of the wire is \( 60 \, \text{cm} \).

Answer 2

Step 1: Given data.
Initial resonating length, l₁ = 90 cm
Initial frequency, f₁ = 400 Hz
New frequency, f₂ = 600 Hz
Tension and mass per unit length remain constant.

Step 2: Relation between frequency and resonating length.
For a sonometer wire under constant tension:
\[ f \propto \frac{1}{l} \] Therefore, \[ \frac{f₁}{f₂} = \frac{l₂}{l₁} \]

Step 3: Substitute the given values.
\[ \frac{400}{600} = \frac{l₂}{90} \] \[ l₂ = 90 \times \frac{400}{600} = 90 \times \frac{2}{3} = 60 \, \text{cm} \]

Step 4: Final Answer.
The new resonating length of the wire is:
\[ \boxed{60 \, \text{cm}} \]

Final Answer: 60 cm