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%

Key concepts

Frequency Calculation

To understand how to calculate the frequency of a vibrating string, we start by looking at the wave equation. This equation shows how the frequency depends on the tension of the wire, its length, and the mass per unit length. Mathematically, the frequency \( f \) is given by the formula:\[f = \dfrac{1}{2L} \sqrt{\dfrac{F}{µ}}\]Here, \( F \) stands for the tension in the wire, \( \mu \) (mu) is the mass per unit length, and \( L \) is the length of the string. All these factors together influence how fast the wire vibrates. The bigger the tension, for instance, the higher the frequency, because the tighter the wire, the quicker it vibrates.
If you're asked to calculate this, make sure you know all these variables. Plug them into the formula, and you can find the frequency of any wire under tension!
Keep in mind that changes in any of these variables can directly affect the frequency and hence produce different sounds or pitches.

Tension and Frequency Relationship

The connection between tension and frequency is crucial when talking about vibrating strings. As mentioned in the wave equation, tension \( F \) directly influences the frequency \( f \). The stronger the tension, the higher the frequency, and vice versa. This is because:

• The wire vibrates faster when pulled tighter.
• The square root relationship shows that frequency doesn't increase linearly with tension but rather with the square root of the tension.

For example, if you increase the tension, the frequency goes up, but not by the full amount of that increase due to the square root factor.
This relationship means that if the tension is reduced by a specific percentage, the frequency changes less dramatically. For a tension change, frequency changes by the square root of that change, explaining why when the tension decreases by \(19\%\), the frequency only reduces by around \(10\%\).
This is because the square root of \(0.81\) (from reducing tension by \(19\%\)) is about \(0.9\), hence reducing \(10\%\) from the original frequency.

Percentage Change in Frequency

Calculating the percentage change in frequency involves understanding how changes in tension affect it according to the wave equation. When the tension of a string changes, its frequency changes too, in a way determined by the square root of the new to old tension ratio.For instance, if the tension decreases by \(19\%\), we compute the new tension as \(0.81F_1\) where \(F_1\) is the initial tension. The frequency change can be calculated using:

• \(f_1\): the initial frequency.
• \(f_2\): the new frequency, which is found from the equation for \(f\) mentioned earlier.

The percentage change is given by:\[\text{Percentage decrease} = \left(1 - \sqrt{0.81}\right) \times 100\%\]This results in approximately \(10\%\) decrease. It illustrates that even a significant change in tension results in a smaller percentage change in frequency due to the square root relation.
Grasping these principles helps us predict how different control factors in string instruments or other tension-driven systems affect sound and frequency, bringing science to help us tune instruments accurately.