Question
13. A string is fixed at both ends. On increasing the tension in the string by 2.5 N , the fundamental frequency is altered in the ratio of \( 3: 2 \). What is the original stretching force? (Ans: 2 N )
Ask by Ellis Washington. in Saudi Arabia
Feb 24,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The original stretching force is 2 N.
Solution
For a string fixed at both ends, the fundamental frequency is given by
f = (1/2L) • √(T/μ)
where T is the tension and μ is the linear mass density.
If the tension is increased from T to (T + 2.5) N, then the frequency becomes
f' = (1/2L) • √((T + 2.5)/μ).
Hence, the ratio of the new frequency to the original frequency is
f'/f = √((T + 2.5)/T).
We are told f'/f = 3/2, so
(3/2) = √((T + 2.5)/T).
Squaring both sides yields
(9/4) = (T + 2.5)/T.
Rearranging, we have
9T = 4(T + 2.5)
9T = 4T + 10
9T - 4T = 10
5T = 10
T = 2 N.
Thus, the original stretching force is 2 N.
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Simplify this solution The Deep Dive
When analyzing a string fixed at both ends, the fundamental frequency \( f \) is related to the tension \( T \) by the equation \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( L \) is the length of the string and \( \mu \) is the mass per unit length. If the tension increases by 2.5 N and the frequency ratio changes to \( \frac{3}{2} \), you can set up the following relationship to solve for the original tension. Using the ratios of the squares of the frequencies gives us an equation involving the original and the new tension. By substituting the known values into the equation, you can arrive at the original stretching force being 2 N, creating a happy ending!
