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# A string with both ends held fixed is vibrating in its

ISBN: 9780321675460 31

## Solution for problem 76P Chapter 15

University Physics | 13th Edition

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University Physics | 13th Edition

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Problem 76P

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 cm from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

Step-by-Step Solution:

Solution 76P Step 1: Suppose, the equation for the standing wave, y (x,t) = 2A sin (kx) sin (t) So, if x =0, sin (kx) = 0 So, the amplitude of the wave will be zero, y (0,t) = 0 Similarly, at y (L,t) = 0 because, the end also will contribute for a node here. Therefore, sin (kL) = 0 -1 Or, kL = sin 0 = n Or, k = n / L So, we can write, y (x,t) = 2A sin [(n/L)x] sin (t) Step 2: Provided, at an antinode, y (x,t) = 0.400 cm = 0.004 m Suppose, at x = L/2, the wave provides an antinode. Therefore, y (x,t) = 2A sin [(n/L)L/2] sin (t) y (x,t) = 2A sin [(n/2)] sin (t) sin (n/2) = - 1 or +1 Provided, the wave is in 3rd harmonics. Therefore, n = 3 Therefore, sin (3/2) = - 1 So, y (x,t) = - 2A sin (t) So, we can write, 0.004 m = - 2A sin (t) 2A = - 0.004 m / sin (t) We can take modulus value for the amplitude. Therefore, y (x,t) = 2A sin [(3/L)x] sin (t) Substituting for 2A from our new expression. y (x,t) = (0.004 m / sin (t)) sin [(3/L)x] sin (t) y (x,t) = 0.004 m sin [(3/L)x] Step 3: Relation between velocity, frequency and wavelength is, v = f Therefore, = v/f Provided, v = 192 m/s and f = 240 Hz = 192 m/s / 240 Hz = 0.8 m Equation for wavelength is, = n / 2L n Therefore, 2L = / n n L = n 2n Provided, n = 3 Therefore, L = 0.80 m / 6 = 0.133 m y (x,t) = 0.004 m sin [(3/0.133 m)x] Step 4: a) i) At x = 40 cm = 0.40 m, y (0.4,t) = 0.004 m sin [(3/0.133) 0.40 m] y (0.4,t) = 0.004 m sin (9) = 0 So it is a node. ii) At x = 20 cm = 0.20 m, y (0.2,t) = 0.004 m sin [(3/0.133) 0.20 m] y (0.2,t) = 0.004 m sin (3×3/2) y (0.2,t) = 0.004 m sin (9/2) sin (9/2) = 1 y (0.2,t) = 0.004 m So it is an antinode. iii) At x = 10 cm = 0.10 m, y (0.1,t) = 0.004 m sin [(3/0.133) 0.10 m] y (0.1,t) = 0.004 m sin (3× ¾ ) = 0 y (0.1,t) = 0.004 m sin (9/4) sin (9/4) = 0.707 y (0.1,t) = 0.004 m × 0.707 = 0.0028 m

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##### ISBN: 9780321675460

This full solution covers the following key subjects: string, its, displacement, amplitude, calculate. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. This textbook survival guide was created for the textbook: University Physics, edition: 13. Since the solution to 76P from 15 chapter was answered, more than 295 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 76P from chapter: 15 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. University Physics was written by and is associated to the ISBN: 9780321675460. The answer to “A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 m/s and a frequency of 240 Hz. The amplitude of the standing wave at an antinode is 0.400 cm. (a) Calculate the amplitude at points on the string a distance of (i) 40.0 cm; (ii) 20.0 cm; and (iii) 10.0 cm from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).” is broken down into a number of easy to follow steps, and 117 words.

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