Problem 57 - University Physics with Modern Physics

Question

CP Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F .\) (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { beat }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 Hz. One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?

Step-by-Step Solution

Answer

The string tension changed by approximately 0.682%.

Step 1: Understand the Frequency Relationship

For a string under tension, the fundamental frequency \( f \) is given by:\[ f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} ,\] where \( L \) is the length of the string, \( F \) is the tension in the string, and \( \mu \) is the mass per unit length. Since the strings are identical and the initial tension is the same, they start with the same frequency \( f_{0} \).

Step 2: Derive Change in Frequency

When the tension \( F \) of one string changes to \( F + \Delta F \), the new frequency \( f' \) for that string becomes:\[ f' = \frac{1}{2L} \sqrt{\frac{F + \Delta F}{\mu}} .\] To simplify, consider small changes, we use a first order Taylor expansion for the square root: \[ \sqrt{F + \Delta F} \approx \sqrt{F} + \frac{1}{2\sqrt{F}} \Delta F . \] Therefore, \( f' \approx f_{0} + \frac{f_{0}}{2F} \Delta F \).

Step 3: Calculate Beat Frequency

The beat frequency \( f_{\text{beat}} \) is the absolute difference between the frequencies of the two strings: \[ f_{ ext{beat}} = |f' - f_{0}| = \left| f_{0} + \frac{f_{0}}{2F} \Delta F - f_{0} \right| = f_{0}\frac{\Delta F}{2F} .\] This establishes the relationship \( f_{\text{beat}} = f_{0} \frac{\Delta F}{2F} \).

Step 4: Use Given Conditions to Find Change in Tension

Given \( f_{\text{beat}} = 1.5 \text{ Hz} \) and \( f_{0} = 440 \text{ Hz} \), solve for \( \frac{\Delta F}{F}\) using the beat frequency formula.\[ 1.5 = 440 \frac{\Delta F}{2F} \Rightarrow \frac{\Delta F}{F} = \frac{2(1.5)}{440} = \frac{3}{440} . \]

Step 5: Calculate the Percentage Change in Tension

Convert the ratio \( \frac{\Delta F}{F} = \frac{3}{440} \) to a percentage.\[ \text{Percentage change} = \left( \frac{3}{440} \right) \times 100\% \approx 0.682\% .\]

Key Concepts

Beat Frequency

Beat frequency is a fascinating concept often observed in musical instruments. When two frequencies move close together, beats are created. This is perceived as a fluctuating volume or 'wave'.
It occurs when two waves of different frequencies interfere. The beat frequency is simply the absolute difference between these two frequencies.

• If you have frequencies of 440 Hz and 441.5 Hz, the beat frequency is 1.5 Hz.
• The equation to calculate beat frequency is: \( f_{\text{beat}} = |f_1 - f_2| \)

Understanding beats helps musicians to tune instruments correctly. When beats are minimized to zero, the two frequencies are in harmony.

Frequency of a String

The frequency at which a string vibrates is vital to the sound it produces. It depends on several factors including the string’s length, tension, and mass per unit length.
The fundamental frequency of a string is given by:

• \( f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \)

Where:

• \( L \) is the length of the string.
• \( F \) is the tension in the string.
• \( \mu \) is the mass per unit length.

The frequency increases if the tension increases or mass per unit length decreases. This relationship allows musicians to alter the pitch of a note by tightening or loosening a string.

Tension of a String

The tension of a string is critical in determining its vibration frequency. Greater tension results in higher pitch, making it essential in musical instrument design.
Tension affects frequency through the equation:

• \( f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \)

Increasing tension increases the sound frequency, making the sound higher. For example, tuning a guitar involves adjusting the tension to create the correct notes. This precise adjustment ensures the string vibrates at the desired frequency, producing the right tone.

Mass Per Unit Length

Mass per unit length, commonly denoted as \( \mu \), plays a significant role in how a string vibrates. It is essentially the mass of a string divided by its length, often expressed in kilograms per meter.

• It impacts the frequency: heavier strings (larger \( \mu \)) vibrate slower, producing lower frequencies.
• Lighter strings are used for higher notes due to their quicker vibrational frequency.

For musicians, selecting the right string with appropriate mass per unit length is crucial for achieving the desired sound. It affects both the tone and playability of a musical instrument.