Question

University Physics I Spring 2019 7. (15 pts) A horizontal string of length L has one end fixed and the other end free to move vertically (but not horizontally). The relationship between L and the wavelength X for standing waves on the string is _2m +1, (1) where m = 0 corresponds to the fundamental, m = 1 to the first overtone, m = 2 to the second overtone, etc. Suppose the wavelength of overtone m is 9 cm and the wavelength of the next higher overtone (m +1) is 7 cm. (a) (5 pts) What is the value of m? Hint: The length of the string remains the same for the two overtones, and m must be a whole number. (b) (5 pts) What is the length of the string? (c) (5 pts) If the tension in the string is 10 N and its mass per length is 0.05 kg/m, what is the frequency of the fundamental? (Answer in Hz.) Hint: Recall that the speed of SH waves on a string with tension T and mass per unit length is v= T/M.

Answer 1

2 +1 = Կեն m+) - 4L - 4L 2mt! 2(mil) +1 2m+ 3 4L dividin Soth Antl. 2m+ 3 2 m + 2m+3 4L OX 7a) Xm=9cm , Amtl = 7cm using X= 4L the dont = +1) + mint 3 dividing both parents 2nti 2m+) I solving for m 1 2m+3 7(2m+3) = 912m+1) 14 m +21 - 18m +9 = 4m = 12 (m=3] b) using Az= 9cm = 42 9 = 44 R = $2cm = 15.75cm c) The speed of the wave => v= Tu ve ry thorong Valo 12 m Now the fundamental frequency f = Ž - f = lore = 3052 2(3)+1 4L 201)+1 20)+1 If=0-673 HZ