Question

A string \(25 \mathrm{~cm}\) long and having a mass of \(2.5 \mathrm{~g}\) is under tension. A pipe closed at one end is \(40 \mathrm{~cm}\) long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second is heard. It is observed that decreasing the tension in the string decreases the beat frequency. The speed of sound in air is \(320 \mathrm{~ms}^{-1}\) The frequency of the string vibrating in its \(1^{\text {st }}\) overtone is \(\ldots \ldots \ldots . \mathrm{Hz}\) (A) 92 (B) 108 (C) 192 (D) 208 .

Short answer

The frequency of the string vibrating in its \(1^{\text{st}}\) overtone is \(208 \mathrm{~Hz}\) (Answer: D).

Step-by-step solution

Understand and write down the given information

From the exercise, we have the following information: Length of the string (L) \(= 25\mathrm{~cm}\) mass of the string (m) \(= 2.5\mathrm{~g}\) Length of the closed pipe \(= 40\mathrm{~cm}\) Speed of sound in air \(= 320\mathrm{~ms}^{-1}\) Beat frequency \(= 8\mathrm{~Hz}\)

Calculate the fundamental frequency of the closed pipe

Since the pipe is closed at one end, its fundamental frequency is given by: \(f_{\text{pipe}} = \frac{v}{4 \times L_{\text{pipe}}}\) Where \(v\) is the speed of sound and \(L_{\text{pipe}}\) is the pipe's length. Plugging in the given values, \(f_{\text{pipe}} = \frac{320\mathrm{~ms}^{-1}}{4\times0.40\mathrm{~m}} = 200\mathrm{~Hz}\)

Calculate the linear density of the vibrating string

To calculate the frequency of the string in its first overtone, we will need its linear density, which is given by: \(\mu = \frac{m}{L}\) Converting the mass and length into SI units, we have: \(m\) \(= 2.5\mathrm{~g} = 0.0025\mathrm{~kg}\) \(L\) \(= 25\mathrm{~cm} = 0.25\mathrm{~m}\) Now, calculating the linear density: \(\mu = \frac{0.0025\mathrm{~kg}}{0.25\mathrm{~m}} = 0.01\mathrm{~kg/m}\)

Calculate the string's first overtone frequency

Now, considering the beat frequency, we have that the string's first overtone frequency can either be 8 Hz higher or lower than the fundamental frequency of the pipe: \(f_1 = f_{\text{pipe}} + 8\mathrm{~Hz} = 200\mathrm{~Hz} + 8\mathrm{~Hz} = 208\mathrm{~Hz}\) \(f_2 = f_{\text{pipe}} - 8\mathrm{~Hz} = 200\mathrm{~Hz} - 8\mathrm{~Hz} = 192\mathrm{~Hz}\) Next, we'll apply the formula for the frequency of the string in the first overtone mode: \(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\) Since decreasing the tension in the string decreases the beat frequency, we should choose \(f_1 = 208 \mathrm{~Hz}\) as the frequency for the first overtone. Therefore, the frequency of the string vibrating in its \(1^{\text{st}}\) overtone is \(208 \mathrm{~Hz}\) (Answer: D).

Key concepts

String Vibration

When a string vibrates, it produces sound waves that we can hear. This is how we get music from instruments like guitars and violins. A string's vibration depends on various factors such as its length, tension, and mass. Often, we're interested in finding the frequency at which the string vibrates, as this determines the pitch of the sound.
For any string fixed at both ends, like those on most musical instruments, the vibration can occur in different modes. The simplest mode is the fundamental frequency, but strings can also vibrate in overtones, which are higher frequency vibrations. In this exercise, the string vibrates in its first overtone. This means the frequency of vibration is twice that of its fundamental frequency.
Mathematically, the frequency of a vibrating string in its first overtone can be calculated using:\( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)
Where:

• \( f \) is the frequency.
• \( T \) is the tension of the string.
• \( \mu \) is the linear density.
• \( L \) is the length of the string.

This calculation is crucial for musicians and engineers to design instruments or systems that have the desired sound characteristics.

Beat Frequency

Beat frequency is a fascinating concept that arises when two sound waves of slightly different frequencies interfere with each other. This interference produces a phenomenon called 'beats,' which is a pulsing or throbbing sound. The beat frequency, specifically, is the difference between the two frequencies.
In the context of this exercise, when the string and the pipe resonate, they produce beats if their frequencies are not equal. If the sound from the vibrating string has a frequency slightly different than the fundamental frequency of the closed pipe, beats can be heard.
Mathematically, the beat frequency can be represented as:\[ f_{\text{beat}} = | f_{1} - f_{2} | \]Where:

• \( f_{1} \) and \( f_{2} \) are the frequencies of the two sound sources.

Understanding beat frequency is essential in tuning musical instruments, as musicians listen for beats to decide if two instruments are in harmony. In this exercise, the beat frequency is given as 8 Hz, which indicates whether the string's vibration matches the pipe's fundamental note or not.

Closed Pipe Resonance

Resonance in a closed pipe is an interesting phenomenon in acoustics, similar to how blowing over a bottle's opening can produce sound. Resonance occurs when an object vibrates at its natural frequency, producing a loud sound. For closed pipes, like the pipe mentioned in the exercise, there is a specific way they resonate.
In a closed-end pipe, only odd harmonics of fundamental frequency are possible, which means the pipe resonates only at odd multiples of its fundamental frequency. This is because a closed pipe has one end closed, forcing the sound wave to produce a node at the closed end and an antinode at the open end. The fundamental frequency \( f_{\text{pipe}} \) is expressed as:\( f_{\text{pipe}} = \frac{v}{4L} \)Where:

• \( v \) is the speed of sound.
• \( L \) is the length of the pipe.

Closed pipe resonance plays a significant role in instruments like clarinets and organs, helping to produce distinct notes. In this problem, the pipe has a fundamental frequency of 200 Hz, which interacts with the vibrating string to produce beats.

Linear Mass Density

Linear mass density is a crucial factor when discussing wave frequency on strings, as it influences how a string vibrates. It is defined as the mass of the string per unit of length. The linear mass density, \( \mu \), can be determined using:\( \mu = \frac{m}{L} \)Where:

• \( m \) is the mass of the string.
• \( L \) is the length of the string.

The linear mass density affects the speed at which waves travel along the string, which in turn influences the frequency. A denser string will generally vibrate at a lower frequency because it requires more energy to move.Understanding linear mass density is vital when designing or adjusting string instruments to ensure they produce the right tones. In our given exercise, with a mass of 2.5 g and a length of 25 cm, the linear mass density was calculated to be 0.01 kg/m, which helped solve for the string's frequency in its overtone.