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Step-by-Step Solution
Step 1: Identify the Known Parameters
• Linear mass density of the wire, \mu = 9.0 \times 10^{-4} \, \text{kg/m}
• Tension in the wire, T = 900 \, \text{N}
• First resonant frequency, f_{1} = 500 \, \text{Hz}
• Next higher resonant frequency, f_{2} = 550 \, \text{Hz}
Step 2: Write Down the Formula for Wave Speed on the String
The wave speed V on a stretched string is given by
V = \sqrt{\frac{T}{\mu}}.
Substituting T = 900 \, \text{N} and \mu = 9.0 \times 10^{-4} \, \text{kg/m} :
V = \sqrt{\frac{900}{9.0 \times 10^{-4}}} = \sqrt{10^6} = 1000 \, \text{m/s}.
Step 3: Express the Resonant Frequencies in Terms of the String Length
The resonant frequencies for a string with fixed ends can be written as
f_{n} = \frac{nV}{2L},
where
• n is an integer (the mode number),
• V is the wave speed,
• L is the length of the string.
Step 4: Relate the Given Frequencies to Consecutive Modes
We are given that the frequencies 500 Hz and 550 Hz are successive resonant frequencies. Therefore, if the first one is f_{n} = 500 \,\text{Hz} , the next one is f_{n+1} = 550 \,\text{Hz} . Mathematically:
f_{n} = \frac{nV}{2L} = 500,
f_{n+1} = \frac{(n+1)V}{2L} = 550.
Step 5: Subtract the Two Equations to Find \frac{V}{2L}
Subtracting the first equation from the second:
f_{n+1} - f_{n} = \frac{(n+1)V}{2L} - \frac{nV}{2L} = \frac{V}{2L}.
Hence,
550 - 500 = \frac{V}{2L} \implies 50 = \frac{V}{2L}.
Step 6: Calculate the Length of the Wire
Since V = 1000 \,\text{m/s} , we get
50 = \frac{1000}{2L} \implies 50 = \frac{1000}{2L}.
Solving for L :
50 = \frac{1000}{2L} \implies 2L = \frac{1000}{50} = 20 \implies L = 10 \,\text{m}.
Step 7: Final Answer
The length of the wire is 10 m.
