Problem 60 - University Physics with Modern Physics

Question

A New Musical Tnstrument. You have designed a new musical instrument of very simple construction. Your design consists of a metal tube with length \(L\) and diameter \(L / 10\) . You have stretched a string of mass per unit length \(\mu\) across the open end of the tube. The other end of the tube is closed. To produce the musical effect you're looking for, you want the frequency of the third-harmonic standing wave on the string to be the same as the fundamental frequency for sound waves in the air column in the tube. The speed of sound waves in this air column is \(v_{v}\) (a) What must be the tension of the string to produce the desired effect? (b) What happens to the sound produced by the instrument if the tension is changed to twice the value calculatod in part (a)?(c) For the tension calculated in part (a), what other harmonics of the string, if any, are in resonance with standing waves in the air column?

Step-by-Step Solution

Answer

(a) T must be \( \left( \frac{v_{v} \cdot 2 \sqrt{\mu}}{6} \right)^2 \). (b) The frequency increases by \( \sqrt{2} \). (c) Only the third harmonic remains in resonance.

Step 1: Identify the frequencies

For the string, the frequency of the third harmonic is given by \( f_3 = \frac{3}{2L} \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension and \( \mu \) is the mass per unit length.For the air column, fundamental frequency is \( f_1 = \frac{v_{v}}{4L} \). Equating \( f_3 \) and \( f_1 \) results in \( \frac{3}{2L} \sqrt{\frac{T}{\mu}} = \frac{v_{v}}{4L} \).

Step 2: Solve for tension

From the equation \( \frac{3}{2L} \sqrt{\frac{T}{\mu}} = \frac{v_{v}}{4L} \), we rearrange and solve for tension: \( \sqrt{T} = \frac{v_{v} \cdot 2 \sqrt{\mu}}{6} \). Squaring both sides gives \( T = \left( \frac{v_{v} \cdot 2 \sqrt{\mu}}{6} \right)^2 \).

Step 3: Identify effect of doubling tension

When tension is doubled to \( 2T \), the new frequency \( f' \) is \( \frac{3}{2L} \sqrt{\frac{2T}{\mu}} = \sqrt{2} \cdot f_3 \). The frequency increases by a factor of \( \sqrt{2} \), which will no longer match the air column's fundamental frequency.

Step 4: Determine harmonics in resonance

We must determine which harmonics satisfy \( f_n = f_1 \) for the tensions calculated. Since the air column is closed at one end, its harmonics are odd multiples: \( f_1, f_3, f_5, \ldots \). Therefore, only the third harmonic on the string is in resonance with the fundamental frequency of the air column for tension \( T \).

Key Concepts

Standing Waves

Standing waves occur when waves traveling in opposite directions superimpose and create a wave that appears to be standing still. They are a fundamental concept in understanding harmonics in musical instruments.
In the case of a string fixed at both ends, standing waves form at specific frequencies where the nodes and anti-nodes align with the physical boundaries of the string.

• Nodes are points of no displacement, meaning the string doesn’t move at these points.
• Anti-nodes are points where the displacement is maximum.

A third-harmonic standing wave on a string has three segments of anti-nodes separated by nodes. It is important because it reveals how energy is distributed along the string, producing different musical notes.

Frequency of Sound Waves

Frequency is a key property of sound waves, dictating the pitch we hear. It’s measured in Hertz (Hz), representing the number of wave cycles passing a point per second. In musical terms, higher frequencies correspond to higher pitches.
For sound waves in a tube closed at one end, the fundamental frequency can be calculated using the formula:
\( f_1 = \frac{v_v}{4L} \)
Here, \( f_1 \) is the fundamental frequency, \( v_v \) is the speed of sound in air, and \( L \) is the length of the tube.

• For the third harmonic on a string, the frequency is given by:
  \( f_3 = \frac{3}{2L} \sqrt{\frac{T}{\mu}} \)
• Matching these frequencies helps create a harmonious sound when both the string and the air column produce waves of the same frequency.

Tension in Strings

Tension in a string is vital to determine the frequency of its vibrations. It's what keeps the string taut and affects the speed of waves traveling through it.
A formula to find the tension needed for a certain frequency is:
\( T = \left( \frac{v_v \cdot 2 \sqrt{\mu}}{6} \right)^2 \)
In this formula, \( T \) represents the tension, \( \mu \) is the mass per unit length of the string, and \( v_v \) is the speed of sound in the air column.

• Increasing the tension increases the frequency, making the pitch higher.
• Decreasing the tension lowers the frequency, making the pitch lower.

Changing tension can disrupt the harmony of the instrument, as it alters the frequency relationship between the string and the air column.

Resonance in Musical Instruments

Resonance is when one system vibrates at the same frequency as another, amplifying the sound. In musical instruments, resonance results in richer and louder tones.
For an air column closed at one end like a tube, the resonance occurs at odd harmonics, such as \( f_1, f_3, f_5 \), etc.

• This means only these frequencies will strongly resonate, amplifying the sound.
• Incorporating resonance ensures the harmonics produced complement each other, leading to a pleasant sound experience.

In the designed instrument, ensuring that the string's third harmonic is in resonance with the air column’s fundamental frequency is critical to achieving the desired musical effect.