A 50.0-gram, 7.20-meter string is subject to a tension force of 45.0 N, with both ends fixed. Adopt a coordinate system in which the string lies along the x-axis, with one end at x = 0 and the other at x = 7.20 m.(a)Determine the positions (x-values) of the nodes and antinodes when the string is exhibiting the third harmonic. (Enter your answers, in m, from smallest to largest.)1st node x = m1st antinode x = m2nd node x = m2nd antinode x = m3rd node x = m3rd antinode x = m4th node x = m(b)At what frequency (in Hz) does this third harmonic vibrate? Hz
Question
A 50.0-gram, 7.20-meter string is subject to a tension force of 45.0 N, with both ends fixed. Adopt a coordinate system in which the string lies along the x-axis, with one end at x = 0 and the other at x = 7.20 m.(a)Determine the positions (x-values) of the nodes and antinodes when the string is exhibiting the third harmonic. (Enter your answers, in m, from smallest to largest.)1st node x = m1st antinode x = m2nd node x = m2nd antinode x = m3rd node x = m3rd antinode x = m4th node x = m(b)At what frequency (in Hz) does this third harmonic vibrate? Hz
Solution
(a) The nodes and antinodes of a standing wave occur at regular intervals along the string. For the third harmonic (n=3), there will be 4 nodes (including the ends of the string) and 3 antinodes. The nodes occur at x = 0, x = L/3, x = 2L/3, and x = L, where L is the length of the string. The antinodes occur at x = L/6, x = L/2, and x = 5L/6.
Given that the length of the string (L) is 7.20 m, we can substitute this value into the above expressions to find the positions of the nodes and antinodes:
1st node x = 0 m 1st antinode x = 7.20 m / 6 = 1.20 m 2nd node x = 7.20 m / 3 = 2.40 m 2nd antinode x = 7.20 m / 2 = 3.60 m 3rd node x = 2 * 7.20 m / 3 = 4.80 m 3rd antinode x = 5 * 7.20 m / 6 = 6.00 m 4th node x = 7.20 m
(b) The frequency of the third harmonic can be found using the formula f = n*v/2L, where v is the speed of the wave on the string. The speed of the wave can be found using the formula v = sqrt(T/μ), where T is the tension in the string and μ is the linear mass density of the string.
First, we need to find the linear mass density of the string: μ = m/L = 50.0 g / 7.20 m = 0.00694 kg/m.
Next, we find the speed of the wave: v = sqrt(45.0 N / 0.00694 kg/m) = 80.6 m/s.
Finally, we can find the frequency of the third harmonic: f = 3 * 80.6 m/s / (2 * 7.20 m) = 16.8 Hz.
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