Question: You are given four tuning forks. The fork with the lowest frequency oscillates at 500 Hz. By striking two tuning forks at a time,you can produce the following beat frequencies, 1, 2, 3, 5, 7, and 8Hz. What are the possible frequencies of the other three forks?(There are two sets of answers)

Short Answer

Expert verified

Answer

The possible frequencies of the other three forks are, f1 = 508 Hz , f2 = 507 Hz ,and f3 = 505 Hz , or, f1 = 508 Hz , f2 = 503 Hz ,and f3 = 501 Hz ,

Step by step solution

01

Step 1: Given

  1. The fork with lowest frequency isf4=500Hz
  2. The beat frequencies are1Hz,2Hz,3Hz,5Hz,7Hz,and8Hz
  3. There are two sets of answer.
02

Determining the conceptBy using the formula for beat frequency fbeat, find the possible frequencies of other three forks.

Formula is as follow:

The beat frequency is,

.fbeat=f1-f2

where, f is frequency.

03

Determining thepossible frequencies of other three forks 

Let,f1f2,f3andf4be the frequencies of the corresponding fork 1, fork 2, fork 3, and fork 4respectively, so that,f1>f2>f3>f4

.

Then, the beat frequency is,

fbeat=f1-f4

Thus, the highest frequency f1 of the corresponding fork is given by,

f1=fbeat+f4f1=8+500f1=508Hz

Similarly, the beat frequency is,

fbeat=f2-f4

Thus, the frequency f4 of the corresponding fork is given by,

f2=fbeat+f4f2=7+500f2=507Hz

Similarly, the beat frequency is,

fbeat=f3-f4

Thus, the frequency f4 of the corresponding fork 3 is given by,

f3=fbeat+f4f3=5+500f3=505Hz

Now, for second set, also f1>f2>f3>f4and

The beat frequency is,

fbeat=f1-f4

Thus, the highest frequency of the corresponding fork is given by,

f1=fbeat+f4f1=8+500f1=508Hz

Similarly, the beat frequency is,

fbeat=f2-f4

Thus, the frequency f2 of the corresponding fork is given by,

role="math" localid="1661435994772" f2=fbeat+f4f2=3+500f2=503Hz

Similarly, the beat frequency is,

fbeat=f3-f4

Thus, the frequency of the corresponding fork 3 is given by,

f3=fbeat+f4f3=1+500f3=501Hz

Hence, the possible frequencies of other three forks are,f1=508Hz,f2=507Hz,andf3=505Hz,or,f1=508Hz,f2=503Hz,andf3=501Hz,.

Thus, by using the formula for beat frequency,fbeatthis question can be solved.

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

Question: Two loud speakers are located 3.35 mapart on an outdoor stage. A listener is18.3m from one andfrom the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range 20Hz to 20 KHz. (a) What is the lowest frequency fmin1that gives minimum signal (destructive interference) at the listener’s location? By what number must fmin1be multiplied to get(b)The second lowest frequencyfmin2that gives minimum signal and(c)The third lowest frequencyfmin3that gives minimum signal ?(d)What is the lowest frequencyfmin1that gives maximum signal (constructive interference) at the listener’s location ? By what number mustfmin1be multiplied to get(e) the second lowest frequencyfmin2that gives maximum signal and(f) the third lowest frequencyfmin3that gives maximum signal?

The A string of a violin is a little too tightly stretched. Beats at 4.00 per second are heard when the string is sounded together with a tuning fork that is oscillating accurately at concert A (440Hz).What is the period of the violin string oscillation?

Question: In Fig. 17-27, pipe Ais made to oscillate in its third harmonicby a small internal sound source. Sound emitted at the right endhappens to resonate four nearby pipes, each with only one openend (they are notdrawn to scale). Pipe Boscillates in its lowestharmonic, pipe Cin its second lowest harmonic, pipe Din itsthird lowest harmonic, and pipe Ein its fourth lowest harmonic.Without computation, rank all five pipes according to theirlength, greatest first. (Hint:Draw the standing waves to scale andthen draw the pipes to scale.)

Four sound waves are to be sent through the same tube of air, in the same direction:

s1(x,t)=(9.00 nm)cos(2πx700πt)s2(x,t)=(9.00 nm)cos(2πx700πt+0.7π)s3(x,t)=(9.00 nm)cos(2πx700πt+π)s4(x,t)=(9.00 nm)cos(2πx700πt+1.7π).

What is the amplitude of the resultant wave? (Hint:Use a phasor diagram to simplify the problem.)

tube of air, transporting energy at the average rate of Pavg,11.In a second experiment, two other sound waves, identical to the first one, are to be sent simultaneously through the tube with a phase differencefof either 0,0.2 wavelength, or 0.5 wavelength between the waves. (a) With only mental calculation, rank those choices of according to the average rate at which the waves will transport energy, greatest first. (b) For the first choice of , what is the average rate in terms of Pavg,11 ?

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