Question

Two sitar strings \(\mathrm{A}\) and \(\mathrm{B}\) playing the note "Dha" are slightly out of time and produce beats of frequency \(5 \mathrm{~Hz}\). The tension of the string B is slightly increased and the beat frequency is found to decrease to \(3 \mathrm{~Hz}\). What is the original frequency of \(\mathrm{B}\) if the frequency of \(\mathrm{A}\) is \(427 \mathrm{~Hz}\) ? (A) 432 (B) 422 (C) 437 (D) 417

Short answer

The original frequency of string B is 422 Hz.

Step-by-step solution

Identifying the given information and variables

From the given problem, we have the following information: - Beat Frequency_1 = 5 Hz (before increasing the tension of string B) - Beat Frequency_2 = 3 Hz (after increasing the tension of string B) - Frequency of String A (f_A) = 427 Hz - Frequency of String B (f_B) =?

Identifying the relation between beat frequencies and frequency tones

Before we proceed, it's essential to know how the beat frequency is related to the two frequencies producing the beats. So, we have: Beat Frequency = |Frequency of String A - Frequency of String B| Now we can write the equation for both given cases. - For the initial case, beat frequency is 5 Hz, so: Beat Frequency_1 = |f_A - f_B| - For the second case, beat frequency is 3 Hz after increasing the tension of string B, so the frequency of string B increases (f_B1 = new frequency of B): Beat Frequency_2 = |f_A - f_B1|

Writing the equation based on given data

Now we substitute the given data into the equations we formulated and solve: 1) Equation for initial case: 5 Hz = |427 Hz - f_B| 2) Equation for the case after increasing the tension of string B: 3 Hz = |427 Hz - f_B1| Since the tension of string B is increased the frequency increases, we have f_B1 = f_B + x, where x is the frequency increase.

Solve the equations

Let's rewrite equation 2 incorporating f_B1 = f_B + x. 3 Hz = |427 Hz - (f_B + x)| We have two unknowns (f_B and x) and two equations, we can solve for f_B. 1) 5 Hz = 427 Hz - f_B OR 5 Hz = f_B - 427 Hz Either f_B = 427 Hz + 5 Hz = 432 Hz (Case 1) or f_B = 427 Hz - 5 Hz = 422 Hz (Case 2) Now let's check, which case holds for the equation with the tension increase: Remember, x is a positive number since the frequency increased when we increased the tension. 2) 3 Hz = 427 Hz - (f_B + x) OR 3 Hz = (f_B + x) - 427 Hz For Case 1: If f_B = 432 Hz, we get x = -2 Hz, which is wrong (x should be positive). For Case 2: If f_B = 422 Hz, we get x = 2 Hz, which is correct (x is positive). So, the original frequency of string B is 422 Hz. Answer: (B) 422

Key concepts

String Tension

String tension plays a pivotal role in determining the pitch or frequency of a stringed instrument. The principle is quite simple: when you increase the tension in a string, the frequency or pitch of the sound it produces increases. Conversely, reducing the tension results in a decrease in frequency.

This is because the vibrating length and the material of the string affect how fast it oscillates. A tightly stretched string vibrates faster, producing a higher frequency. This is fundamentally due to the wave equation on a string, which tells us that frequency is proportional to the square root of tension. Some key points about string tension:

• High tension leads to high frequency.
• Low tension results in low frequency.
• The change in tension affects how sound waves travel through the string.

This concept is critical when tuning instruments, as musicians adjust the tension to achieve the desired pitch. Understanding string tension is a starting point for grasping the relationship between physical properties and sound.

Frequency Change

Frequency change in musical strings is directly tied to variations in tension and other physical characteristics such as length and mass per unit length. When the tension changes, the frequency changes proportionally, following the wave equation. In this exercise, both strings A and B produce a beat frequency, a phenomenon that arises when two close but unequal frequencies play together. Initially, string B's tension is increased, causing its frequency to rise. As a result, the difference between the frequencies of strings A and B, or the beat frequency, decreases from 5 Hz to 3 Hz.

Here's how it works:

• Beat frequency is the absolute difference between two frequencies.
• Increased tension on string B raises its frequency, illustrating the interplay between tension and frequency change.

This concept highlights how manipulating string tension can precisely control the sounds produced, enabling musicians to refine their instruments' tones by minute adjustments.

Sound Waves

Sound waves are the vibrations that travel through air (or another medium) when a string is plucked, struck, or bowed. These waves are detected by our ears and interpreted as sound. In the context of string instruments, sound waves are generated by the oscillation of strings. Every sound wave has its unique properties, such as frequency, which we perceive as pitch, and amplitude, which determines loudness. In this exercise, we focus on the frequency aspect because it directly relates to the pitch of the musical note produced by the strings. Here are some insights on sound waves:

• Sound waves travel through various media, not only air.
• Frequency changes affect pitch in sound waves.
• Amplitude changes affect the loudness of the sound.

Understanding sound waves is crucial for interpreting how instruments produce different sounds. This knowledge helps explain why a slight adjustment in tension can lead to recognizable harmonic differences in the sound waves emitted by musical instruments.