Solution Found!
Answer: A string with both ends held fixed is vibrating in
Chapter 15, Problem 76P(choose chapter or problem)
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).
Questions & Answers
QUESTION:
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).
ANSWER:Solution 76P
Step 1:
Suppose, the equation for the standing wave,
y (x,t) = 2A sin (kx) sin (𝜔t)
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
Or, kL = sin-1 0 = n𝜋
Or, k = n𝜋 / L
So, we can write, y (x,t) = 2A sin [(n𝜋/L)x] sin (𝜔t)