helppp please- A horizontal string of length 1.5 meters vibrates with a wave velocity of 1320 m/s ...?

Question

A horizontal string of length 1.5 meters vibrates with a wave velocity of 1320 m/s at its fundamental frequency.
a)what is the fundamental frequency?
b)What is the frequency of the fourth overtone, and how many nodes and antinodes will it have?
c)In terms of interference, to what do the nodes and antinodes correspond?
d)How many grams must the string be to have a tension of 10^5N?

absent all week,and I am so lost,pleaseeeee help it will be GREATLY appreciated!

Answer 1

a) String vibration modes have nodes at both ends (the ends of the string are fixed). So the fundamental waveform is one which has a single antinode in between: this represents going from 0 to 1 and 0 again in the sin function (for example), so it's half a wavelength. So the fundamental wavelength is 3m and the fundamental frequency is 1320/3 = 440 Hz.

b) Each overtone will add one node and antinode to the waveform, i.e. half of a full wave. So we get the following results, where l is the length of the string and λ is the wavelength:
Fundamental: 2 nodes, 1 antinode, l = λ/2
First overtone: 3 nodes, 2 antinodes, l = λ
Second overtone: 4 nodes, 3 antinodes, l = 3λ/2
Third overtone: 5 nodes, 4 antinodes, l = 2λ
Fourth overtone: 6 nodes, 5 antinodes, l = 5λ/2.
This gives us λ = 2l/5 = 0.6 m, so the frequency is 1320/0.6 = 2200 Hz.

c) Nodes represent destructive interference, antinodes represent constructive interference.

d) c = sqrt(T/(m/L)), so c^2 = TL/m, so m = TL/c^2
= 10^5 (1.5) / (1320^2)
= 0.086 g.