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 required is approximately 56.753 N.

Step-by-step solution

Convert string length to meters

The given length of the string is 35.0 cm, but we need it to be in meters to be consistent with units in the formula. To convert to meters, divide by 100: $$ L = \frac{35.0 \ \mathrm{cm}}{100} = 0.35 \ \mathrm{m} $$

Write down the known values

Using the converted length, we now list the known values: - Length of string, \(L = 0.35 \ \mathrm{m}\) - Mass per unit length of the string, \(\mu = 5.51*10^{-4} \ \mathrm{kg/m}\) - Fundamental frequency, \(f = 660 \ \mathrm{Hz}\)

Rearrange the formula to solve for tension

Rearrange the formula for the fundamental frequency to solve for the tension \(T\): $$ T = \mu \left(2L f\right)^2 $$

Plug in values and solve for tension

Now, we can plug in the known values into the formula and solve for \(T\): $$ T = \left(5.51 * 10^{-4} \ \mathrm{kg/m}\right) \left(2 * 0.35 \ \mathrm{m} * 660 \ \mathrm{Hz}\right)^2 $$ Calculate the tension: $$ T \approx 56.753 \ \mathrm{N} $$ The tension required to make the string vibrate at the fundamental frequency of 660 Hz is approximately 56.753 N.

Key concepts

String Tension Calculation

The ability to calculate the tension in a vibrating string is crucial for understanding a wide range of musical and physical phenomena. When a string is plucked or struck, the tension contributes to its vibrational properties, which in turn determine the pitch of the sound produced. To find out the necessary tension to achieve a particular frequency, we use the formula connecting tension (T), mass per unit length (\textbf{\textmu}), and frequency (f):

\textbf{Formula for Tension:} \[ T = \textbf{\textmu} (2Lf)^2 \]
Surprisingly, this formula shows that the tension increases with the square of the frequency, highlighting how dramatically tension must change to reach higher pitches. In our exercise, the result was approximately 56.753 Newtons (N). Understanding string tension is important not only for musicians tuning their instruments but also for engineers and physicists dealing with materials and resonances.

Mass Per Unit Length

Mass per unit length (often denoted as \textbf{\textmu}) is a fundamental property of strings that directly affects their vibrational characteristics. It is defined as the mass of the string divided by its length:

\textbf{Definition:} \[ \textbf{\textmu} = \frac{\text{Mass}}{\text{Length}} \]
This property is key to determining the string’s inertia and resistance to acceleration during vibration. A higher mass per unit length implies a slower response to force, resulting in a lower pitch, while a lower mass per unit length allows for higher frequencies. In the given exercise, we used the value of \( 5.51 \times 10^{-4} \text{ kg/m} \) to calculate the tension needed for a specific vibrational frequency. When designers craft musical instruments, such as guitars or pianos, they carefully consider the mass per unit length to ensure the instrument can produce the desired range of notes.

Harmonics in Physics

Harmonics, or overtones, are an essential concept in physics, explaining the phenomenon of multiples of a fundamental frequency. When a string vibrates, it doesn't just produce one single note but also generates a series of higher-pitched sounds at frequencies that are integer multiples of the fundamental frequency (f).

\textbf{Harmonic Series:} \[ f_n = n \times f \]
Here, \( f_n \) represents the frequency of the nth harmonic, and n is an integer. The first harmonic (n=1) is the fundamental frequency, the second harmonic (n=2) is twice the fundamental, and so forth. This series enriches the sound produced by musical instruments and is also a cornerstone in understanding complex waveforms. The understanding of harmonics not only aids in music theory and instrumentation but is also pivotal in areas such as acoustics, electronics, and even telecommunications.