A string of a violin produces 2 beats per second when sounded along with a standard fork of frequency \(400 . \mathrm{Hz}\). The beat frequency increases when the string is tightened. a) What was the frequency of the violin at first? b) What should be done to tune the violin?

Short Answer

Expert verified
Answer: The initial frequency of the violin string was 398 Hz. To tune the violin, the string should be tightened until its frequency reaches 400 Hz to match the frequency of the tuning fork.

Step by step solution

01

Understand the beat frequency

Beat frequency is the difference between the frequencies of two waves. In this case, the beat frequency is created when the violin string is played along with a standard tuning fork. The beat frequency is given as 2 beats per second (Hz).
02

Use the beat frequency formula to find the initial frequency

The formula for beat frequency is: Beat frequency = |f1 - f2| where f1 and f2 are the frequencies of the two waves. In this case, we are given the beat frequency (2 Hz) and the frequency of the tuning fork (400 Hz). Let the initial frequency of the violin string be f_violin. Then we have: 2 = |f_violin - 400|
03

Solve for the initial frequency of the violin string

We have two possibilities for the initial frequency: 1. f_violin - 400 = 2, which implies f_violin = 402 Hz 2. 400 - f_violin = 2, which implies f_violin = 398 Hz Since we are given that the beat frequency increases when the string is tightened, we can infer that the initial frequency of the violin string must be 398 Hz (lower than the tuning fork). Therefore, the initial frequency of the violin string is 398 Hz. a) The frequency of the violin at first was 398 Hz.
04

Determine the action needed to tune the violin

We are given that the beat frequency increases when the string is tightened, which means the frequency of the violin string also increases as it approaches the frequency of the tuning fork (400 Hz). To tune the violin, we need to adjust the string to match the frequency of the tuning fork. b) To tune the violin, we should tighten the string until the frequency of the violin string reaches 400 Hz and matches the frequency of the tuning fork. At this point, there will be no beats as both the sources have the same frequency.

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

Two vehicles carrying speakers that produce a tone of frequency \(1000.0 \mathrm{~Hz}\) are moving directly toward each other. Vehicle \(\mathrm{A}\) is moving at \(10.00 \mathrm{~m} / \mathrm{s}\) and vehicle \(\mathrm{B}\) is moving at \(20.00 \mathrm{~m} / \mathrm{s}\). Assume the speed of sound in air is \(343.0 \mathrm{~m} / \mathrm{s}\), and find the frequencies that the driver of each vehicle hears.

A classic demonstration of the physics of sound involves an alarm clock in a bell vacuum jar. The demonstration starts with air in the vacuum jar at normal atmospheric pressure and then the jar is evacuated to lower and lower pressures. Describe the expected outcome.

A policeman with a very good ear and a good understanding of the Doppler effect stands on the shoulder of a freeway assisting a crew in a 40 -mph work zone. He notices a car approaching that is honking its horn. As the car gets closer, the policeman hears the sound of the horn as a distinct \(\mathrm{B} 4\) tone \((494 \mathrm{~Hz}) .\) The instant the car passes by, he hears the sound as a distinct \(\mathrm{A} 4\) tone \((440 \mathrm{~Hz}) .\) He immediately jumps on his motorcycle, stops the car, and gives the motorist a speeding ticket. Explain his reasoning.

Two trains are traveling toward each other in still air at \(25.0 \mathrm{~m} / \mathrm{s}\) relative to the ground. One train is blowing a whistle at \(300 .\) Hz. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). a) What frequency is heard by a man on the ground facing the whistle-blowing train? b) What frequency is heard by a man on the other train?

A train whistle emits a sound at a frequency \(f=3000 .\) Hz when stationary. You are standing near the tracks when the train goes by at a speed of \(v=30.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in the frequency \((|\Delta f|)\) of the whistle as the train passes? (Assume that the speed of sound is \(v=343 \mathrm{~m} / \mathrm{s}\).)

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