Question

A string is \(35.0 \mathrm{~cm}\) long and has a mass per unit length of \(5.51 \cdot 10^{-4} \mathrm{~kg} / \mathrm{m}\). What tension must be applied to the string so that it vibrates at the fundamental frequency of \(660 \mathrm{Hz?}\)

Short answer

Answer: The tension that must be applied to the string is approximately 56.38 N.

Step-by-step solution

Ensure all values are in the correct units

Since the length of the string, \(L\) is given in cm, we need to convert it into meters. The mass per unit length is given in kg/m, which is already in the correct units. \(L = 35.0 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.35 \mathrm{~m}\) Now we have all the values we need: Length of the string, L = 0.35 m Mass per unit length, μ = \(5.51 \times 10^{-4} \mathrm{~kg/m}\) Fundamental frequency, \(f_1 = 660 \mathrm{~Hz}\)

Solve for tension T using the formula

We're solving for \(T\) using the formula: $$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$ First, we'll isolate T by squaring both sides of the equation and then multiply by the denominator: $$T = \mu(2Lf_1)^2$$ Next, we'll plug in the values we have for L, μ, and \(f_1\): $$T = (5.51 \times 10^{-4} \mathrm{~kg/m})\left(2\times(0.35 \mathrm{~m})\times(660 \mathrm{~Hz})\right)^2$$

Calculate the tension T

Now, we can perform the calculations: $$T = (5.51 \times 10^{-4} \mathrm{~kg/m}) \times (4\times0.35 \mathrm{~m}\times660 \mathrm{~Hz})^2$$ $$T ≈ 56.38 \mathrm{~N}$$ This gives the tension in the string that must be applied for it to vibrate at the fundamental frequency of 660 Hz.

Key concepts

Fundamental Frequency

The fundamental frequency of a string refers to the lowest frequency at which the string can naturally vibrate. When you pluck or strike a string, this is the main frequency you hear. The fundamental frequency is determined by several factors, including the length of the string, the tension in the string, and the mass per unit length. For a string fixed at both ends, the fundamental frequency is calculated using the equation:\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]Where: - \( f_1 \) is the fundamental frequency - \( L \) is the length of the string - \( T \) is the tension in the string - \( \mu \) is the mass per unit length of the string This equation shows the important relationship between these variables. A higher tension or a lower mass per unit length will increase the fundamental frequency, making the pitch higher.

String Tension

Tension in a string is the force applied to stretch the string and is crucial for determining the pitch of the sound produced. Tension affects the speed of the wave moving through the string, which consequently affects the frequency. Increasing the tension results in a higher wave speed, leading to a higher frequency, while reducing the tension leads to a lower frequency. To calculate the correct tension for the desired fundamental frequency, you can rearrange the fundamental frequency equation to solve for tension:\[ T = \mu (2Lf_1)^2 \]Using this formula, you can easily adjust the tension of a string to achieve the desired frequency. Keep in mind that excessive tension can lead to overstretching and potential breaking of the string.

Mass per Unit Length

The mass per unit length, often represented by the symbol \( \mu \), is a measure of how much mass is distributed across a given length of the string. It is directly tied to the material and thickness of the string. This factor is crucial in determining the string's frequency characteristics. A string with a higher mass per unit length will have a lower frequency because the wave moves slower compared to a lighter string. Conversely, a lighter string (lower mass per unit length) will produce a higher frequency. You can use the relation:\[ \mu = \frac{m}{L} \]Where \( m \) is the total mass of the string and \( L \) is its length. This measure allows you to adjust the frequency by opting for different materials or thicknesses when designing or selecting a string for a specific application.

Vibrations of a String

The vibrations of a string produce sound through the creation of standing waves. When a string is displaced, waves travel along the length of the string, reflecting back and forth as they hit the fixed ends. These waves interfere in such a way as to create points along the string, called nodes, where no movement occurs, and antinodes, where the maximum movement occurs. These standing waves dictate the frequencies produced by the string. The fundamental mode, or the first harmonic, is the simplest form of such a vibration with one antinode in the middle of the string and nodes at each end. Higher harmonics can occur, with increased numbers of nodes and antinodes, though the fundamental frequency is always the most dominant. Vibrations are influenced by tension, mass per unit length, and other factors such as the surrounding medium and temperature.