Question

The tension in a wire is decreased by \(19 \%\), then the percentage decrease in frequency will be....... (A) \(19 \%\) (B) \(10 \%\) (C) \(0.19 \%\) (D) None of these

Short answer

The percentage decrease in frequency when the tension in a wire is decreased by \(19\%\) is approximately \(10\%\).

Step-by-step solution

Recall the relationship between tension and frequency in a wire

According to the wave equation for a stretched wire, the frequency of a vibrating wire depends on the tension (F), the mass per unit length (µ), and the length of the wire (L). Mathematically, this can be represented as: \(f = \dfrac{1}{2L} \sqrt{\dfrac{F}{µ}}\) Where: - f is the frequency of the wire - F is the tension in the wire - µ is the mass per unit length - L is the length of the wire

Determine the new frequency

Let the initial frequency be \(f_1\) and the initial tension be \(F_1\). When the tension decreases by 19%, the new tension \(F_2\) can be found as: \(F_2 = F_1 - 0.19F_1 = 0.81F_1\) Now, let's find the new frequency \(f_2\) when the tension is F_2: \(f_2 = \dfrac{1}{2L} \sqrt{\dfrac{F_2}{µ}} = \dfrac{1}{2L} \sqrt{\dfrac{0.81F_1}{µ}}\)

Calculate the percentage decrease in frequency

To find the percentage decrease in frequency, we first need to find the ratio of the new frequency to the initial frequency: Percentage decrease = \(\dfrac{f_1 - f_2}{f_1} \times 100\%\) Substitute \(f_1\) and \(f_2\) from the earlier equations: Percentage decrease = \(\dfrac{\dfrac{1}{2L} \sqrt{\dfrac{F_1}{µ}} - \dfrac{1}{2L} \sqrt{\dfrac{0.81F_1}{µ}}}{\dfrac{1}{2L}\sqrt{\dfrac{F_1}{µ}}} \times 100\% \) Simplify the equation: Percentage decrease = \(\dfrac{\sqrt{\dfrac{F_1}{µ}} - \sqrt{\dfrac{0.81F_1}{µ}}}{\sqrt{\dfrac{F_1}{µ}}} \times 100\% \) Percentage decrease = \(1 - \sqrt{0.81} \times 100\% \) Percentage decrease ≈ 10% Therefore, the correct option is: (B) 10%