Question

A column of air of length \(50 \mathrm{~cm}\) resonates with a stretched string of length \(40 \mathrm{~cm}\). The length of the same air column which will resonate with \(60 \mathrm{~cm}\) of the same string at the same tension is \(\ldots \ldots \ldots\) (A) \(100 \mathrm{~cm}\) (B) \(75 \mathrm{~cm}\) (C) \(50 \mathrm{~cm}\) (D) \(25 \mathrm{~cm}\)

Short answer

The length of the air column that would resonate with a 60cm long string is approximately 48cm. Considering the multiple choice options, the answer is approximately (C) \(50 \mathrm{~cm}\).

Step-by-step solution

Understand the situation

Here, we have a situation in which an air column resontes with a stretched string. The initial length of the air column is 50cm, and the string's length is 40cm. We need to find the length of the air column that would resonate with a 60cm long string, maintaining the same tension.

Write the formula for frequency of the string

The frequency of the string (f) can be determined using the following formula: \(f = \frac{v}{2L}\) Where \(v\) is the velocity of the wave, and \(L\) is the length of the string.

Determine the relationship between the initial conditions and the final conditions

In this case, we know that for the column of air to resonate with the string, the frequency of the string must match the natural frequency of the air column. Since the tension in the string is constant, we can deduce that the velocity of the waves in the string is also constant.

Write an equation to represent the relationship between the initial and final conditions

Using the relationship between the frequency, length, and velocity of a wave, we can write the equation as follows: \( \frac{v}{2(40)} = \frac{v}{2(50)} \times \frac{x}{60} \) Where \(x\) is the length of the new air column.

Solve for x

Now, we need to solve the equation for x. First, we can simplify it: \( \frac{v}{80} = \frac{v}{100} \times \frac{x}{60} \) Now, we can remove the constant velocity (\(v\)) from both sides and solve for \(x\): \( 80 = \frac{100x}{60} \) \( 80 \times 60 = 100x \) \( x = \frac{80 \times 60}{100} \) \( x= 48 \) The length of the air column that would resonate with a 60cm long string is approximately 48cm. Since this is an approximation, considering the multiple choice options, the answer would be: (C) \(50 \mathrm{~cm}\) [approximation of the result]

Key concepts

Frequency of Waves

The frequency of a wave is a fundamental concept in physics. It's essentially how often the wave oscillates over a fixed period of time. Imagine you’re watching a wave in the ocean; the frequency would be how many waves pass a point in a certain amount of time. For a musical string instrument, this concept translates to how many times the string vibrates. This, in turn, determines the pitch of the sound you hear. In physics terms, frequency is measured in Hertz (Hz), which counts the number of complete cycles per second.
For strings, frequency is determined by several factors: the length of the string, the tension of the string, and the wave velocity. If you change one of these factors, you alter the frequency. In the given exercise, the constant tension maintains the velocity of the waves in the string, which directly influences the frequency since that part of the equation remains unchanged.
The key equation relevant here is: \( f = \frac{v}{2L} \), where

• \(f\) = frequency
• \(v\) = wave velocity
• \(L\) = length of the string

Understanding how these components interact helps solve problems related to resonance, such as the one in the exercise.

Resonance and Harmonics

Resonance is a phenomenon where an object or system vibrates at a larger amplitude at certain frequencies, known as resonant frequencies. In everyday terms, it’s like finding the sweet spot where maximum vibration occurs. In the context of musical instruments and air columns, resonance manifests as loud, sustained notes. For the air column to resonate with a string, their frequencies must match.
In the exercise, resonance occurs in an air column due to waves generated by a vibrating string. Harmonics, on the other hand, are integral multiples of fundamental frequency, and they indicate the different resonance possibilities. For instance, the first harmonic is the fundamental frequency itself — often the simplest wave pattern — while the second, third, and others are more complex wave patterns fitting within the same space. This concept is critical as it explains why different lengths of string and air column can resonate with each other.
When adjusting string length, as in the exercise's scenario, understanding harmonics helps determine the equivalent length of the air column to maintain resonance. Resonance can create impressive effects, like sustaining sounds in musical instruments or even catastrophic consequences in engineering if not properly managed.

Wave Velocity in Strings

Wave velocity in strings is another crucial aspect when discussing frequency and resonance. In simple terms, wave velocity refers to how fast the wave travels along the string. This speed can be influenced by the tension in the string and its mass per unit length. The velocity is mathematically represented as \( v = \sqrt{\frac{T}{\mu}} \), where

• \(T\) = tension in the string
• \(\mu\) = linear mass density or mass per unit length

In your main exercise, the tension is held constant, which means the wave velocity remains unchanged regardless of the string length. Hence, as you change the length, you indirectly change the frequency to maintain the resonant condition with the air column. By understanding the wave velocity’s consistency, it becomes easier to navigate how frequency adjusts with length changes. This makes calculations, as shown in the step-by-step solution, more straightforward. This consistency helps in keeping a stable pitch and ensuring that instruments sound the way they're meant to.