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Chapter 10: Problem 1461

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

Expert verified
The original frequency of string B is 422 Hz.

Step by step solution

01

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) =?
02

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|
03

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.
04

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

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Most popular questions from this chapter

What should be the speed of a source of sound moving towards a stationary listener, so that the frequency of sound heard by the listener is double the frequency of sound produced by the source? \\{Speed of sound wave is \(\mathrm{v}\\}\) (A) \(\mathrm{v}\) (B) \(2 \mathrm{v}\) (C) \(\mathrm{v} / 2\) (D) \(\mathrm{v} / 4\)

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