/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q18DQ By touching a string lightly at ... [FREE SOLUTION] | 91影视

91影视

By touching a string lightly at its center while bowing, a violinist can produce a note exactly one octave above the note to which the string is tuned鈥攖hat is, a note with exactly twice the frequency. Why is this possible?

Short Answer

Expert verified

The frequency of the vibration gets double when the finger is touched at the middle vibrating string of the guitar.

Step by step solution

01

Concept of frequency of the string

The frequency of a string is directly proportional to the square root of its tension.

The frequency of the string is given by

f=12LT

Here, L is the length of the string, T is the tension, is the linear mass density.

02

Relation between the frequencies of the string.

The frequency of the string vibration is inversely proportional to the length of the string.

If the length of the string is short then the frequency is high. When the violinist touches center of the string, a node is produced at that point and the length of the vibrating element is half of the actual element.

The formula to calculate the frequency is,

f=12LF 鈥... (1)

For center position, the length becomes half, Substitutefor in equation (1)

f'=12L2F 鈥.. (2)

Here. f' is the frequency of vibration when plucked at the center of the string.

Divide equations (2) and (1)

f'f=1LF12Ff'=2f

Therefore, the frequency of the vibration gets double when the finger touched at the middle vibrating string of guitar.

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 91影视!

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 you stretch a rubber band and pluck it, you hear a (somewhat) musical tone. How does the frequency of this tone change as you stretch the rubber band further? (Try it!) Does this agree with Eq. (15.35) for a string fixed at both ends? Explain.

In case 1, a source of sound approaches a stationary observer at speed u. In case 2, the observer moves toward the stationary source at the same speed u. If the source is always producing the same frequency sound, will the observer hear the same frequency in both cases, since the relative speed is the same each time? Why or why not?

BIO The Human Voice. The human vocal tract is a pipe that extends about 17 cm from the lips to the vocal folds (also called 鈥渧ocal cords鈥) near the middle of your throat. The vocal folds behave rather like the reed of a clarinet, and the vocal tract acts like a stopped pipe. Estimate the first three standing-wave frequencies of the vocal tract. Use v = 344 m/s. (The answers are only an estimate, since the position of lips and tongue affects the motion of air in the vocal tract.)

A railroad train is traveling at 25m/s in still air. The frequency of the note emitted by the locomotive whistle is 400Hz. What is the wavelength of the sound waves (a) in front of the locomotive and (b) behind the locomotive? What is the frequency of the sound heard by a stationary listener (c) in front of the locomotive and (d) behind the locomotive?

Longitudinal Waves on a Spring. A long spring such as a SlinkyTMis often used to demonstrate longitudinal waves. (a) Show that if a spring that obeys Hooke鈥檚 law has mass m, length L, and force constant k鈥, the speed of longitudinal waves on the spring is v=Lk'm

(see Section 16.2). (b) Evaluate v for a spring with m = 0.250 kg, L = 2.00 m, and k鈥 = 1.50 N/m.

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.