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Chapter 10: Problem 1490

The frequency of tuning fork \(\mathrm{A}\) is \(2 \%\) more than the frequency of a standard fork. Frequency of tuning fork \(B\) is \(3 \%\) less than the frequency of the standard fork. If 6 beats per second are heard when the two forks \(\mathrm{A}\) and \(\mathrm{B}\) are excited, then frequency of \(A\) is \(\ldots \ldots \ldots . H z\) (A) 120 (B) \(122.4\) (C) \(116.4\) (D) 130

Short Answer

Expert verified
The frequency of tuning fork A is \(122.4\, Hz\), which corresponds to option (B).

Step by step solution

01

Setting up the equations

Let the standard frequency be S. According to the problem, frequency of A is 2% more than S, so: \(A = S + 0.02S = 1.02S\) Similarly, frequency of B is 3% less than S, so: \(B = S - 0.03S = 0.97S\) We also know that the absolute difference between the frequencies of A and B is 6 Hz. \(|A - B| = 6\)
02

Solving for the standard frequency (S)

We know that \(A = 1.02S\) and \(B = 0.97S\). Substitute these into the equation \(|A - B| = 6\), we get: \(|1.02S - 0.97S| = 6\) Solve for S: \(|0.05S| = 6\) \(S = \frac{6}{0.05}\) \(S = 120\)
03

Find the frequency of A

We now know the standard frequency is 120 Hz. To find the frequency of A, we substitute the value of S into the equation: \(A = 1.02S\) \(A = 1.02 \times 120\) \(A = 122.4\, Hz\) Therefore, the frequency of tuning fork A is 122.4 Hz, which corresponds to option (B).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Beat Frequency
When two sound waves of different frequencies are played together, they interfere with each other (a process called superposition). This results in a fluctuation or variation in loudness, which we perceive as beats.
The beat frequency is simply the number of these fluctuations or cycles in loudness per second. It can be calculated using the formula:
  • Beat Frequency = \(|f_1 - f_2|\)
Here, \(f_1\) and \(f_2\) are the frequencies of the two sound sources. In our problem, these are tuning forks A and B.
It's stated that 6 beats per second are heard when tuning forks A and B are excited past.
This implies that the beat frequency is 6 Hz, which means \(|A - B| = 6\). This helps us in setting up an equation to solve for the unknown frequencies.
Percent Change in Frequency
Understanding percent change is crucial in problems dealing with frequency adjustments. A percent change describes how much a value increases or decreases in terms of a percentage of the original value.
For tuning fork A, its frequency is 2% more than a standard frequency S. This means:
  • Tuning Fork A Frequency = Standard Frequency + 2% of Standard Frequency.
  • \(A = S + 0.02S = 1.02S\)
Similarly, for tuning fork B, which is 3% less than S:
  • Tuning Fork B Frequency = Standard Frequency - 3% of Standard Frequency.
  • \(B = S - 0.03S = 0.97S\)
By understanding these percentage changes, we can set up the equations necessary to solve for the unknown frequencies. It's a straightforward application of what percent increase and percent decrease mean.
Equation Solving
To find the unknown frequencies, setting up the correct equations is key. Based on the problem, we have two expressions:
  • \(A = 1.02S\)
  • \(B = 0.97S\)
And we also know that \(|A - B| = 6\). The equation becomes:
  • \(|1.02S - 0.97S| = 6\)
  • This simplifies to \(|0.05S| = 6\)
We can solve for S (the standard frequency) by dividing both sides by 0.05:
  • \(S = \frac{6}{0.05}\)
  • \(S = 120\)
Once we have S, calculating the frequency of tuning fork A is straightforward:
  • \(A = 1.02 \times 120\)
  • \(A = 122.4\, Hz\)
This logical progression from setting up an equation through solving it is a clear example of how mathematical reasoning can lead to a solution. It's all about breaking down the problem into manageable steps and dealing with them one at a time.

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Most popular questions from this chapter

The bob of a simple pendulum having length ' \(\ell\) ' is displaced from its equilibrium position by an angle of \(\theta\) and released. If the velocity of the bob, while passing through its equilibrium position is \(\mathrm{v}\), then \(\mathrm{v}=\ldots \ldots \ldots\) (A) \(\sqrt{\\{2 g \ell(1-\cos \theta)\\}}\) (B) \(\sqrt{\\{2 g \ell(1+\sin \theta)\\}}\) (C) \(\sqrt{\\{2 g \ell(1-\sin \theta)\\}}\) (D) \(\sqrt{\\{2 g \ell(1+\cos \theta)\\}}\)

A metal wire having linear mass density \(10 \mathrm{~g} / \mathrm{m}\) is passed over two supports separated by a distance of \(1 \mathrm{~m}\). The wire is kept in tension by suspending a \(10 \mathrm{~kg}\) mass. The mid point of the wire passes through a magnetic field provided by magnets and an a. c. supply having frequency \(\mathrm{n}\) is passed through the wire. If the wire starts vibrating with its resonant frequency, what is the frequency of a. c. supply? (A) \(50 \mathrm{~Hz}\) (B) \(100 \mathrm{~Hz}\) (C) \(200 \mathrm{~Hz}\) (D) \(25 \mathrm{~Hz}\)

A simple pendulum is executing S.H.M. around point \(\mathrm{O}\) between the end points \(B\) and \(C\) with a periodic time of \(6 \mathrm{~s}\). If the distance between \(\mathrm{B}\) and \(\mathrm{C}\) is \(20 \mathrm{~cm}\) then in what time will the bob move from \(C\) to \(D\) ? Point \(D\) is at the mid-point of \(C\) and \(\mathrm{O}\). (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)

As shown in figure, a spring attached to the ground vertically has a horizontal massless plate with a \(2 \mathrm{~kg}\) mass in it. When the spring (massless) is pressed slightly and released, the \(2 \mathrm{~kg}\) mass, starts executing S.H.M. The force constant of the spring is \(200 \mathrm{Nm}^{-1}\). For what minimum value of amplitude, will the mass loose contact with the plate? (Take \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(10.0 \mathrm{~cm}\) (B) \(8.0 \mathrm{~cm}\) (C) \(4.0 \mathrm{~cm}\) (D) For any value less than \(12.0 \mathrm{~cm}\).

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of \((\mathrm{g} / 3)\), then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)
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