A cello string has a fundamental frequency of \(65.40 \mathrm{Hz}\) What beat frequency is heard when this cello string is bowed at the same time as a violin string with frequency of \(196.0 \mathrm{Hz} ?\) [Hint: The beats occur between the third harmonic of the cello string and the fundamental of the violin. \(]\)

Short Answer

Expert verified
Answer: The beat frequency between the cello string's third harmonic and the violin string's fundamental frequency is \(0.20 \mathrm{Hz}\).

Step by step solution

01

Calculate the Cello String's Third Harmonic Frequency

We are given the fundamental frequency of the cello string as \(65.40 \mathrm{Hz}\). The third harmonic frequency is three times the fundamental frequency: $$ f_{cello\_harmonic\_3} = 3 \times 65.40 \mathrm{Hz} $$ Find the cello's third harmonic by multiplying the given fundamental frequency by three.
02

Calculate the Beat Frequency

Beat frequency is the absolute difference between the frequencies of two waves. In this case, we're comparing the cello's third harmonic and the violin's fundamental frequencies: $$ f_{beat} = |f_{violin\_fundamental} - f_{cello\_harmonic\_3}| $$ We already have the violin's fundamental frequency given as \(196.0 \mathrm{Hz}\) from the problem. After calculating the cello's third harmonic, we can subtract the cello's value from the violin's to find their absolute difference.
03

Solve for the Beat Frequency

First, use the result from step 1 to find the third harmonic frequency of the cello string: $$ f_{cello\_harmonic\_3} = 3 \times 65.40 \mathrm{Hz} = 196.20 \mathrm{Hz} $$ Now, substitute the values of the violin's fundamental frequency and the cello's third harmonic frequency into the beat frequency formula: $$ f_{beat} = |196.0 \mathrm{Hz} - 196.20 \mathrm{Hz}| $$ Calculate the absolute difference between the two frequencies to find the beat frequency.
04

Final Answer

The beat frequency between the cello string's third harmonic and the violin string's fundamental frequency is: $$ f_{beat} = |196.0 \mathrm{Hz} - 196.20 \mathrm{Hz}| = 0.20 \mathrm{Hz} $$ So, the beat frequency heard is \(0.20 \mathrm{Hz}\).

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.

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

The Vespertilionidae family of bats detect the distance to an object by timing how long it takes for an emitted signal to reflect off the object and return. Typically they emit sound pulses 3 ms long and 70 ms apart while cruising. (a) If an echo is heard 60 ms later $\left(v_{\text {sound }}=331 \mathrm{m} / \mathrm{s}\right),\( how far away is the object? (b) When an object is only \)30 \mathrm{cm}$ away, how long will it be before the echo is heard? (c) Will the bat be able to detect this echo?
In an experiment to measure the speed of sound in air, standing waves are set up in a narrow pipe open at both ends using a speaker driven at \(702 Hz\). The length of the pipe is \(2.0 \mathrm{m} .\) What is the air temperature inside the pipe (assumed reasonably near room temperature, \(20^{\circ} \mathrm{C}\) to \(\left.35^{\circ} \mathrm{C}\right) ?\) [Hint: The standing wave is not necessarily the fundamental.]
An intensity level change of \(+1.00 \mathrm{dB}\) corresponds to what percentage change in intensity?
According to a treasure map, a treasure lies at a depth of 40.0 fathoms on the ocean floor due east from the lighthouse. The treasure hunters use sonar to find where the depth is 40.0 fathoms as they head cast from the lighthouse. What is the elapsed time between an emitted pulse and the return of its echo at the correct depth if the water temperature is $\left.25^{\circ} \mathrm{C} ? \text { [Hint: One fathom is } 1.83 \mathrm{m} .\right]$
A violin is tuned by adjusting the tension in the strings. Brian's A string is tuned to a slightly lower frequency than Jennifer's, which is correctly tuncd to \(440.0 \mathrm{Hz}\) (a) What is the frequency of Brian's string if beats of \(2.0 \mathrm{Hz}\) are heard when the two bow the strings together? (b) Does Brian need to tighten or loosen his A string to get in tune with Jennifer? Explain.
See all solutions

Recommended explanations on Physics Textbooks

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