Question:
A fundamental harmonic standing wave is generated in a string fixed at both ends. If the tension in the string is increased by \(21\%\), the percentage change in the fundamental frequency will be:
A fundamental harmonic standing wave is generated in a string fixed at both ends. If the tension in the string is increased by \(21\%\), the percentage change in the fundamental frequency will be:
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For large percentage increases (like \(21\%\)), do not use the small-error differential approximation (\(\Delta f / f \approx \frac{1}{2} \Delta T / T\)), as that shortcut is strictly meant for variations under \(5\%\). Instead, seek out perfect squares hidden in the decimals (like \(1.21 = 1.1^2\) or \(1.44 = 1.2^2\)) to quickly pull values out of the radical!
Updated On: May 25, 2026
- \(10\%\)
- \(21\%\)
- \(11\%\)
- \(10.5\%\)
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The Correct Option is A
Solution and Explanation
Concept:
The fundamental frequency (\(f\)) of a transverse standing wave in a stretched string fixed at both ends of length \(L\) depends on the speed of the wave (\(v\)) along the string:
\[
f = \frac{v}{2L}
\]
The wave speed \(v\) is determined by the tension (\(T\)) in the string and its linear mass density (\(\mu\), mass per unit length) via the relationship:
\[
v = \sqrt{\frac{T}{\mu}}
\]
Combining these two equations yields the direct dependency relation for the fundamental frequency:
\[
f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \quad \Rightarrow \quad f \propto \sqrt{T}
\]
Since the length \(L\) and linear mass density \(\mu\) remain perfectly constant during the tension adjustment, the frequency varies directly with the square root of the tension.
Step 1: Establish the algebraic variables for the initial state.
Let the initial tension in the string be \(T_1 = T\), and the corresponding initial fundamental frequency be: \[ f_1 = k\sqrt{T} \quad \cdots (1) \] where \(k\) is a constant of proportionality encompassing \(\frac{1}{2L\sqrt{\mu}}\).
Step 2: Determine the new tension in terms of the initial state.
The problem states that the tension is increased by \(21\%\). Calculating the final tension \(T_2\): \[ T_2 = T + 21\% \text{ of } T = T + 0.21T = 1.21T \]
Step 3: Calculate the final fundamental frequency \(f_2\).
Substitute the updated tension value into our proportionality relation: \[ f_2 = k\sqrt{1.21T} = \sqrt{1.21} \cdot k\sqrt{T} \] Since \(\sqrt{1.21} = 1.1\), substitute equation (1) back into the expression: \[ f_2 = 1.1 f_1 \quad \cdots (2) \]
Step 4: Compute the percentage change in the fundamental frequency.
The fractional change in frequency is defined as: \[ \text{Fractional Change} = \frac{f_2 - f_1}{f_1} \] Substituting our value from equation (2): \[ \text{Fractional Change} = \frac{1.1f_1 - f_1}{f_1} = \frac{0.1f_1}{f_1} = 0.1 \] Convert this value to a percentage by multiplying by \(100\): \[ \text{Percentage Change} = 0.1 \times 100 = 10\% \]
Let the initial tension in the string be \(T_1 = T\), and the corresponding initial fundamental frequency be: \[ f_1 = k\sqrt{T} \quad \cdots (1) \] where \(k\) is a constant of proportionality encompassing \(\frac{1}{2L\sqrt{\mu}}\).
Step 2: Determine the new tension in terms of the initial state.
The problem states that the tension is increased by \(21\%\). Calculating the final tension \(T_2\): \[ T_2 = T + 21\% \text{ of } T = T + 0.21T = 1.21T \]
Step 3: Calculate the final fundamental frequency \(f_2\).
Substitute the updated tension value into our proportionality relation: \[ f_2 = k\sqrt{1.21T} = \sqrt{1.21} \cdot k\sqrt{T} \] Since \(\sqrt{1.21} = 1.1\), substitute equation (1) back into the expression: \[ f_2 = 1.1 f_1 \quad \cdots (2) \]
Step 4: Compute the percentage change in the fundamental frequency.
The fractional change in frequency is defined as: \[ \text{Fractional Change} = \frac{f_2 - f_1}{f_1} \] Substituting our value from equation (2): \[ \text{Fractional Change} = \frac{1.1f_1 - f_1}{f_1} = \frac{0.1f_1}{f_1} = 0.1 \] Convert this value to a percentage by multiplying by \(100\): \[ \text{Percentage Change} = 0.1 \times 100 = 10\% \]
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