Question

A guitar string is \(0.750 \mathrm{~m}\) long and has a mass of \(5.00 \mathrm{~g}\). The string is tuned to \(\mathrm{E}(660 \mathrm{~Hz})\) when it vibrates at its fundamental frequency. What is the required tension on the string? a) \(2.90 \cdot 10^{3} \mathrm{~N}\) b) \(4.84 \cdot 10^{3} \mathrm{~N}\) c) \(6.53 \cdot 10^{3} \mathrm{~N}\) d) \(8.11 \cdot 10^{3} \mathrm{~N}\) e) \(1.23 \cdot 10^{4} \mathrm{~N}\)

Short answer

a) 5.28 x 10^3 N b) 6.18 x 10^3 N c) 6.53 x 10^3 N d) 7.46 x 10^3 N Answer: c) 6.53 x 10^3 N

Step-by-step solution

Convert Units

First, we need to convert the given information to the appropriate units. The string's length is given in meters, which is already the SI unit for length. However, we need to convert the string's mass from grams to kilograms (SI unit for mass): $$ m = 5.00\,\text{g} = 0.00500\,\text{kg} $$

Calculate Linear Density

We need to find the linear density (\(\mu\)) of the string, which is the mass per unit length. To find this, we simply divide the mass of the string (in kilograms) by its length (in meters): $$ \mu = \frac{m}{L} = \frac{0.00500\,\text{kg}}{0.750\,\text{m}} = 0.00667\,\text{kg/m} $$

Use the Fundamental Frequency Formula

We will now use the formula for the fundamental frequency of a vibrating string, which relates the frequency (\(f\)), string length (\(L\)), linear density (\(\mu\)), and tension (\(T\)): $$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $$ We will solve this equation for the tension \(T\), given the string's fundamental frequency \(f = 660\,\text{Hz}\): $$ T = \mu \cdot (2Lf)^2 $$

Calculate Tension

Now we will substitute the values of \(f\), \(L\), and \(\mu\) into the equation and solve for \(T\): $$ T = 0.00667\,\text{kg/m} \cdot (2(0.750\,\text{m})(660\,\text{Hz}))^2 $$ $$ T = 0.00667\,\text{kg/m} \cdot (993.75\,\text{Hz})^2 $$ $$ T \approx 6.53 \times 10^3\,\text{N} $$ The required tension on the string is approximately \(6.53 \times 10^3\,\text{N}\), which corresponds to the choice (c).

Key concepts

Linear Density

Understanding the concept of linear density is crucial for solving problems involving the fundamental frequency of a vibrating string. Linear density, represented by the Greek letter \(\mu\), refers to the amount of mass per unit length of a string or any one-dimensional object. In simple terms, it tells you how much the string 'weighs' per meter.

To calculate linear density, you divide the total mass of the string by its entire length, as demonstrated in our exercise. The unit of linear density is kilograms per meter (\(kg/m\)). A higher linear density means the string is heavier for its length, which generally translates to a slower vibration and a lower fundamental frequency.

For the guitar string in the exercise, we calculated \(\mu = \frac{0.00500\,\text{kg}}{0.750\,\text{m}} = 0.00667\,\text{kg/m}\), which then played a pivotal role in determining the required tension to achieve the correct pitch.

String Tension

String tension is another key factor in the physics of musical instruments. The tension applied to a string affects its vibration and, consequently, the pitch of the sound it produces. Tension is a force measured in newtons (\(N\)), and in the context of a string, it's the stretching force that keeps the string taut.

From the formula \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\), it's clear that tension has a direct relationship with the frequency of the vibration. If the tension increases, the frequency of the vibrating string also increases, leading to a higher pitch. Conversely, lowering the tension decreases the frequency and the pitch. Solving the given equation from the exercise for tension \(T\) and substituting the known values of frequency \(f\), string length \(L\), and linear density \(\mu\), allowed us to find the exact tension needed to tune the guitar string to the note E at 660 Hz.

Harmonic Waves

Now, let's dive into the concept of harmonic waves which governs the behavior of vibrating strings and wind instruments. When a string vibrates, it does so in a standing wave pattern, holding certain fixed points, known as nodes, and oscillating at points in between, known as antinodes.

The fundamental frequency is the simplest form of vibration for a string, with only two nodes at its ends and one antinode in the middle. This is the lowest frequency at which a string can vibrate, and all other frequencies at which the string can naturally vibrate are called harmonics or overtones. These higher frequencies are integer multiples of the fundamental frequency.

In essence, harmonic waves are key to understanding musical tones, as they produce the rich sounds we associate with instruments. The frequency of these harmonic waves is determined by the length, tension, and linear density of the string, creating the musical notes we hear.