Log out Go to App
Go to App

Learning Materials

Features

Discover

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).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If the velocity of sound wave in humid air is \(\mathrm{v}_{\mathrm{m}}\) and that in dry air is \(\mathrm{v}_{\mathrm{d}}\), then \(\ldots \ldots\) (A) \(\mathrm{v}_{\mathrm{m}}>\mathrm{v}_{\mathrm{d}}\) (B) \(\mathrm{v}_{\mathrm{m}}<\mathrm{v}_{\mathrm{d}}\) (C) \(\mathrm{v}_{\mathrm{m}}=\mathrm{v}_{\mathrm{d}}\) \((\mathrm{D}) \mathrm{v}_{\mathrm{m}} \gg \mathrm{v}_{\mathrm{d}}\)

Which of the following functions represents a travelling wave? (A) \((\mathrm{x}-\mathrm{vt})^{2}\) (B) in \((\mathrm{x}+\mathrm{vt})\) (C) \(\mathrm{e}^{-(\mathrm{x}+\mathrm{vt}) 2}\) (D) \(\\{1 /(\mathrm{x}+\mathrm{vt})\\}\)

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

The displacement of a S.H.O. is given by the equation \(\mathrm{x}=\mathrm{A}\) \(\cos \\{\omega t+(\pi / 8)\\}\). At what time will it attain maximum velocity? (A) \((3 \pi / 8 \omega)\) (B) \((8 \pi / 3 \omega)\) (C) \((3 \pi / 16 \omega)\) (D) \((\pi / 16 \pi)\).

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The maximum acceleration of the particle is $\ldots \ldots . . \mathrm{cm} / \mathrm{s}^{2}$. (A) 4 (B) 12 (C) 20 (D) 28
See all solutions

Recommended explanations on English Textbooks

Rhetoric

Read Explanation

Email

Read Explanation

History of English Language

Read Explanation

Textual Analysis

Read Explanation

Prosody

Read Explanation

Text Comparison

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free