# Physics Superposition and Standing Waves (ch. 18)

• Keithkent09
In summary, the tension must vibrate at 155 Hz in order to form the standing wave pattern shown in (b)
Keithkent09

## Homework Statement

A 6.70 kg object hangs in equilibrium from a string with a total length of L = 5.30 m and a linear mass density of µ = 0.00100 kg/m. The string is wrapped around two light frictionless pulleys that are separated by a distance of d = 2.00 m. (visual is attached) The tension is 41.6 N
At what frequency must the string between the pulleys vibrate in order to form the standing wave pattern shown in (b)? (Visual is attached)

## Homework Equations

f=(n/2L)*sqrt(T/u)

## The Attempt at a Solution

I already figured out the tension so I plugged in the relevant values and what I thought the value for n might be but nothing I tried worked.

#### Attachments

• p14-59.gif
2.7 KB · Views: 473
I get around 150 Hz, is this what you got?

The correct answer is 155 and I got 150 one of the times I worked the problem. What did you use for n?

I used n=3 from the photo. I assume your tension computation was correct. You may want to dbl check to see if it is off a bit.
EDIT seems ok, I get 153hz.

Last edited:
Even though the distance over the waves is 2 meters you still set the n to be equal to 3? Are you using the same equation that I used above? The tension is correct that much I do know.

Keithkent09 said:
Even though the distance over the waves is 2 meters you still set the n to be equal to 3? Are you using the same equation that I used above? The tension is correct that much I do know.
Yea, I checked it and get 153Hz. N is the wave number right? So that's what I used as it looks like 1.5 lambda stretched across the 2 meters or am I missing something?

Yeah I follow that, what did you do with the 1.5 lambda over two meters though. I thought the n=3 was just inserted straight into the equation.

Keithkent09 said:
Yeah I follow that, what did you do with the 1.5 lambda over two meters though. I thought the n=3 was just inserted straight into the equation.

Which is what I did, just taking the cue for what harmonic from the pic.

The tension should be 41.27397 N. Using that tension, I got 152.36979 Hz as the final answer.

Note that 150 N, 152 N, 153 N, and 155 N are the same if we ignore the least significant digit.

Don't worry about it; you did it right.

When I plug three in for n i get 57.5.
For me the worked out equation looks like this:
f=(3/2L)*(sqrt(41.3/.001)) with L equaling 5.3. What did you do differently?
When I got 150 earlier I had n equaling a much larger value.
Sorry it is taking me so long to get this.

Keithkent09 said:
When I plug three in for n i get 57.5.
For me the worked out equation looks like this:
f=(3/2L)*(sqrt(41.3/.001)) with L equaling 5.3. What did you do differently?
When I got 150 earlier I had n equaling a much larger value.
Sorry it is taking me so long to get this.

I took L=2.0m that is the wave is only present in the top segment and stops at the pullies which I took to be points of forced zero amplitude boundary condition.

Oh, I was using L=5.3 instead of just the distance that really mattered as 2. Thank you very much I got the answer now.

## 1. What is the principle of superposition in physics?

The principle of superposition states that when two or more waves of the same type overlap, the resulting wave is the sum of the individual waves. This means that the displacement of the medium at any point is the algebraic sum of the individual displacements caused by each wave.

## 2. How does superposition apply to standing waves?

In standing waves, the principle of superposition still applies. The standing wave is formed by the interference between two waves traveling in opposite directions. The amplitude of the standing wave is the result of the superposition of the amplitudes of the two waves.

## 3. What is the difference between a node and an antinode in standing waves?

A node is a point in a standing wave where the displacement of the medium is always zero. On the other hand, an antinode is a point where the displacement of the medium is at its maximum. In other words, a node is a point of destructive interference while an antinode is a point of constructive interference.

## 4. How do you calculate the wavelength of a standing wave?

The wavelength of a standing wave can be calculated by dividing the distance between two consecutive nodes or antinodes by the number of nodes or antinodes. This is because the wavelength of a standing wave is equal to twice the distance between consecutive nodes or antinodes.

## 5. How does the frequency of a standing wave relate to its harmonics?

The frequency of a standing wave is directly proportional to the number of nodes or antinodes. This means that as the number of nodes or antinodes increases, the frequency of the standing wave also increases. These different frequencies correspond to different harmonics of the standing wave.

• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
6K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
15
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
2K