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

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 $60 \, \text{cm}$.