Question

Example 18.7 The Mistuned Piano Strings Two identical piano strings of length 0.775 m are each tuned exactly to 400 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLVE IT Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats will be heard Categorize We must combine our understanding of the waves under boundary conditions model for strings with our new knowledge of beats. Set up a ratio of the fundamental frequencies of the two strings: Substitute for the wave speeds on the strings: Incorporate that the tension in one string is 1.0% larger than the other; that is, T2 = 1.010T1 - V 1.0907. – 1.05 Solve for the frequency of the tightened string: f2 = 1.005f1 = 1.005(400) Hz = 402 Hz Find the beat frequency: fear = 402 Hz - 400 Hz = 2 Hz Finalize Notice that a 1.0% mistuning in tension leads to an easily audible beat frequency. A piano tuner can use beats to tune a stringed instrument by "beating" a note against a reference tone of known frequency. The tuner can then adjust the string tension until the frequency of the sound it emits equals the frequency of the reference tone. The tuner does so by tightening or loosening the string until the beats produced by it and the reference source become too infrequent to notice. MASTER IT HINTS: GETTING STARTED I'M STUCK! Another two identical piano strings are now used. The first one has a frequency (fundamental) at 255 Hz. When the two are being struck, beats are heard with the period between maxima measured at 0.61 s. From these given information, what are the two possible values for the tension in the second string? Give your answer in terms of percentage difference to the tension of the first string. The lower possible tension is 1.04 X % lower than T. The higher possible tension is 1.05 1x % higher than T. Need Help? Read It

Answer 1

The solution of the problems is given below

Fayet, Date f = 255 Hz f = 255+ + } beat time pod to= 0.615 beat frequency - TB •64 H2 as to << t using Bionomif approx. - Toate 2 ) - Fi - - Text of Chenen bir tension I-Ii x 100 z 1 2 3 Тg 1 2x 100 0.61×255 - t 1.28 % :: 1.28 1. for both the parts da 620m L=860 m first min. at pour D. Maxime at a ie the sources are in Sane phase.

Page No. Date at point D. ie path diff for waves reaching from pont A and B = d. AD-BD = 1 ✓ Ltd - dh a x=2[5 Etaž - dom] Substituting the values da a 6202 +86 0² - 620 x za 110602- 6207 - 2x440.2 x 2 880.4m a 880 m