# Normal mode of string in third harmonic

• jimbo71
In summary: So, in summary, the frequency of the wave and the speed of the wave can be used to calculate the time it takes for the string to go from its largest upward displacement to its largest downward displacement at an antinode.
jimbo71

## Homework Statement

A string that is fixed at both ends is vibrating in the third harmonic. the wave has a speed of 186m/s and a frequency of 225Hz. the amplitude of the string at an antinode is 0.0037m.

How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point?

## Homework Equations

y(x,t)=(Aswsin(kx)(sin(omega*t))

## The Attempt at a Solution

I think there is a node at x=0 and the third harmonic means there are three antinodes between x=0 and x=L. Therefore the antinodes are located at x=1/6L,1/2L,5/6L and the nodes are located at x=0,2/6L,4/6L,L. I set 0.0037m=y(1/6L,t)=(Aswsin(kx))(sin(omega*t)) and solve for t where
L=186/225*3/2
Asw=0.0037
k=(2pi*225)/186
x=1/6L
omega=2pi*225
I found the first antinode to occur at t=1/900s
However when I solve for t with x=2/6L I cannot calculate t because it requires me to divide by zero because sin(k*2/6L)=0
what am i doing wrong?

This is just a thought that might simplify things a lot, the frequency of the wave describes how many times the wave passes a point per second, in this case an antinode. Also, each time the wave passes an antinode it has completed one full cycle from peak to trough and back to peak again.

Your approach to finding the time it takes for the string to go from its largest upward displacement to its largest downward displacement at the antinode is correct. However, the issue with dividing by zero is due to a mistake in your calculation of the second antinode location. The second antinode should be located at x=3/6L, not 2/6L. This means that your calculation for t with x=2/6L should be t=1/450s, not t=1/900s. This correction should allow you to continue with your solution and find the correct time for the string to go from its largest upward displacement to its largest downward displacement at the antinode. Keep up the good work in solving these types of problems!

## 1. What is the normal mode of a string in third harmonic?

The normal mode of a string in third harmonic refers to the vibration pattern of a string when it is divided into three equal segments and oscillates at a frequency that is three times the fundamental frequency.

## 2. How is the normal mode of a string in third harmonic different from the fundamental mode?

The normal mode of a string in third harmonic is characterized by three antinodes and two nodes, whereas the fundamental mode has only one antinode and no nodes. Additionally, the frequency of the third harmonic is three times higher than the fundamental frequency.

## 3. What is the significance of the third harmonic in string vibrations?

The third harmonic is important in string vibrations because it is the first overtone of the fundamental frequency. It is also the first mode that has nodes at both ends of the string, which allows for more complex standing wave patterns.

## 4. How does the string length affect the normal mode of the third harmonic?

The string length does not affect the normal mode of the third harmonic. As long as the string is divided into three equal segments, the frequency and vibration pattern of the third harmonic will remain the same. However, the overall frequency of the string will change depending on its length.

## 5. Can there be higher harmonics in string vibrations?

Yes, there can be higher harmonics in string vibrations. Higher harmonics refer to vibration patterns that have more than three antinodes, with frequencies that are multiples of the fundamental frequency. These harmonics become increasingly complex and are not as commonly observed as the first few harmonics.

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