Question

A wire having a linear mass density \( 9.0 \times 10^{-4} \, kg/m \) is stretched between two rigid supports with a tension of \( 900 \, N \). The wire resonates at a frequency of \( 500 \, Hz \). The next higher frequency at which the same wire resonates is \( 550 \, Hz \). The length of the wire is \(\dots\dots\dots\) m.

Hint:

For any wave system where frequencies are harmonically related (like a string fixed at both ends), the fundamental frequency is always the greatest common divisor of all allowed frequencies.

Answer 1

Correct Answer: 10

Step 1: Understanding the Concept:
For a string fixed at both ends, the resonance frequencies are integral multiples of the fundamental frequency (\( f_n = n \cdot f_1 \)). The difference between two successive resonance frequencies is equal to the fundamental frequency.
Step 2: Key Formula or Approach:
1. Fundamental Frequency: \( f_1 = f_{n+1} - f_n = \frac{v}{2L} \).
2. Wave Speed on String: \( v = \sqrt{\frac{T}{\mu}} \).
Step 3: Detailed Explanation:
1. Find Fundamental Frequency:
\[ f_1 = 550 \, Hz - 500 \, Hz = 50 \, Hz \]
2. Calculate Wave Speed (\( v \)):
Given \( T = 900 \, N \) and \( \mu = 9.0 \times 10^{-4} \, kg/m \).
\[ v = \sqrt{\frac{900}{9 \times 10^{-4}}} = \sqrt{\frac{100}{10^{-4}}} = \sqrt{10^6} = 1000 \, ms^{-1} \]
3. Calculate Length (L):
\[ f_1 = \frac{v}{2L} \implies 50 = \frac{1000}{2L} \]
\[ 100L = 1000 \]
\[ L = 10 \, m \]
Step 4: Final Answer:
The length of the wire is 10 m.