One of the harmonic frequencies for a particular string under tension is 325 Hz.The next higher harmonic frequency is 390 Hz. What harmonic frequency is next higher after the harmonic frequency 195 Hz?

Short Answer

Expert verified

The harmonic frequency that is next higher after the harmonic frequency 195 Hz is 260 Hz.

Step by step solution

01

Given data

Under the tension one of the harmonic frequency for a particular string isf1=325Hz and next higher frequency isf2=390Hz

One of the lower harmonic frequencies isf'1=195Hz

02

Understanding the concept of resonant frequency

The harmonic frequencies are integer multiples of the lowest harmonic frequency. Therefore, the difference between two successive frequencies is the fundamental frequency. Adding the fundamental frequency to the given frequency, we can find the successive higher frequency after it.

03

Calculation for the harmonic frequency

The resonant frequency of the string is given as:

f'=nf

Where, n = 1,2,3 and f is the fundamental frequency.

Hence, the fundamental frequency is 390 - 325 = 65 Hz

The frequency next higher after the given frequency = the given frequency + the fundamental frequency

Therefore, the frequency next higher after the harmonic frequency 195 Hz is

195+65=260 Hz

Hence, the harmonic frequency next higher after the harmonic frequency 195 Hz is 260 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 type of rubber band used inside some baseballs and golf balls obeys Hooke’s law over a wide range of elongation of the band. A segment of this material has an un-stretched length land a mass m. When a force Fis applied, the band stretches an additional lengthl. (a) What is the speed (in terms of m, l, and the spring constant k) of transverse waves on this stretched rubber band? (b) Using your answer to (a), show that the time required for a transverse pulse to travel the length of the rubber band is proportional to 1/l if role="math" localid="1660986246683" ll and is constant if ll.

Figure 16-46 shows transverse accelerationayversus time tof the point on a string at x=0, as a wave in the form ofy(x,t)=ymsin(kx-ωt+ϕ)passes through that point. The scale of the vertical axis is set byas=400m/s2. What isϕ? (Caution:A calculator does not always give the proper inverse trig function, so check your answer by substituting it and an assumed value ofωintoy(x,t)and then plotting the function.

A transverse sinusoidal wave is moving along string in the positive direction of an xaxis with a speed of 80 m/s. At t = 0, the string particle atx = 0 has a transverse displacement of 4.0 cmfrom its equilibrium position and is not moving. The maximum transverse speed of the string particle at x = 0is 16 m/s. (a)What is the frequency of the wave? (b)What is the wavelength of the wave?If y(x,t)=ymsin(kx±ωt+ϕ)is the form of the wave equation, (a)What isym, (b)What is k, (c)What is ω, (d)What is, and (e)What is the correct choice of sign in front ofω?

For a particular transverse standing wave on a long string, one of an antinodes is at x = 0and an adjacent node is at x = 0.10 m. The displacement y(t)of the string particle at x = 0is shown in Fig.16-40, where the scale of y theaxis is set by ys=4.0cm. When t = 0.50 s, What is the displacement of the string particle at (a) x = 0.20 mand x = 0.30 m (b) x = 0.30 m? What is the transverse velocity of the string particle at x = 0.20 mat (c) t = 0.50 sand (d) t = 0.1 s ? (e) Sketch the standing wave atfor the range x = 0to x = 0.40 m.

The equation of a transverse wave on a string isy=(2.0mm)sin[20m-1x-600s-1t] . The tension in the string is 15 N . (a)What is the wave speed? (b)Find the linear density of this string in grams per meter.

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