ps01soln - PHYSICS 218 SOLUTION TO HW 1 Created: August 28,...

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Unformatted text preview: PHYSICS 218 SOLUTION TO HW 1 Created: August 28, 2004 7:39am Last updated: September 5, 2004 5:35pm 1. (a) The frequency of the vibration of the string can be expressed as f = 1 r o o , (1) where is the wavelength of the standing wave formed on the string, o is the tension applied on the string and o is the mass per unit length. We see that the frequency of vibration is proportional to the squareroot of the tension in the string, f o . So if the musician needs to increase the frequency of vibration from 435 Hz to 440 Hz, she has to increase the tension in the string. (b) From equation (1), we have f c f i = r c i . Therefore, we find c i = f c f i 2 = 440 435 2 1 . 023 . (2) 2. (a) We assume that the fundamental harmonic has the dominant contribution to the vibration of the string. Lets look at a string with fixed ends boundary conditions. If ` is the length of the string between the two fixed ends, then the wavelength is given by the expression n 2 = ` , where n = 1 , 2 , 3 , . . . . For fundamental harmonics, we set n = 1. So from equation (1), we have f = 1 2 ` r o o . (3) On the other hand, we also know that the tension and stress are related by o = S A , where A is the cross sectional area of the string, and the mass per unit length and density are related by o = % A . Therefore, we have o o = S % . Substituting this expression in (3), we have ` = 1 2 f s S % . (4) If the frequency is to be 261.6 Hz, we find that ` = 1 2 261 . 6 s 8 10 8 8 10 3 = 0 . 604 m . (b) Given the diameter of the string, D , is 1 mm, the tension in the string is o = 1 4 S D 2 = 628 N . Compared to a person of 135 ` b 61 kg, which is about 598 N, the tension in the string is higher than the weight. 1 PHYSICS 218 SOLUTION TO HW 1 (c) Using equation (2), we find that (low A) = 27 . 5 261 . 6 2 628 N = 6 . 94 N , (high C) = 4186 261 . 6 2 628 N = 1 . 61 10 5 N . (d) If f 1 and f 2 are respectively the frequencies of vibration of two strings and A 1 and A 2 are respectively their cross sectional area, then from equation (3), f = 1 2 ` r o % A , we see that if the length and tension of the strings are unchanged, then we have f 1 f 2 = r A 2 A 1 = D 2 D 1 , where D 1 and D 2 are the respective diameters of the two string. Therefore, we find that D (low A) = 261 . 6 27 . 5 1 mm = 9 . 51 mm , D (high C) = 261 . 6 4186 1 mm = 0 . 06 mm . (e) In part (2c), the high C string will almost certainly break since the stress would be about a factor of 250 larger than the nominal maximum stress. A compromise is utilized in real pianos; the diameter of the string is slightly smaller than that of middle C (approximately 0.8 mm) and the string is very short to make the frequency high.0....
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ps01soln - PHYSICS 218 SOLUTION TO HW 1 Created: August 28,...

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