How to maintain the same stationary wave pattern?

In summary: So changing the density or tension will change the wave speed, but why is this relevant to the answer?Ok, thanks. So changing the density or tension will change the wave speed, but why is this relevant to the answer?The answer suggests that we increase the speed of wave instead. Is it done to match the increase in frequency? But even so, why is B wrong?The answer suggests that we increase the speed of wave instead. Is it done to match the increase in frequency? But even so, why is B wrong?
  • #1
toforfiltum
341
4

Homework Statement


upload_2015-7-8_1-24-40.png


Homework Equations


not sure

The Attempt at a Solution


The answer is C, though I answered B. I don't understand why C is the answer.I thought that to maintain the same stationary wave pattern, I must maintain the speed of the stationary wave, and one way to do this is by reducing the wavelength of the wave. Instead, the answer suggests that we increase the speed of wave instead. Is it done to match the increase in frequency? But even so, why is B wrong?
 
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  • #2
In order to have a standing wave, the string length needs to be a half-integer multiple of the wavelength. You therefore cannot change the wavelength. How would you achieve this when the frequency changes? (Hint: How does frequency relate to wave speed?)
 
  • #3
Orodruin said:
In order to have a standing wave, the string length needs to be a half-integer multiple of the wavelength. You therefore cannot change the wavelength. How would you achieve this when the frequency changes? (Hint: How does frequency relate to wave speed?)
Ah, I forgot the fact that the wavelength of a stationary wave cannot be simply changed. Anyway, how do you increase the speed of wave on the string?
 
  • #4
toforfiltum said:
Ah, I forgot the fact that the wavelength of a stationary wave cannot be simply changed. Anyway, how do you increase the speed of wave on the string?

Have you seen the derivation of the wave equation for a string? If not you should look it up, it will tell you the wave velocity in terms of the density and string tension. Changing either will change the wave velocity.
 
  • #5
Orodruin said:
Have you seen the derivation of the wave equation for a string? If not you should look it up, it will tell you the wave velocity in terms of the density and string tension. Changing either will change the wave velocity.
Ok, thanks.
 

FAQ: How to maintain the same stationary wave pattern?

1. How do I create a stationary wave pattern?

To create a stationary wave pattern, you will need a medium (such as a rope or a string), a source of vibration (such as a hand or a machine), and two fixed points to attach the medium. The vibration from the source will cause the medium to form a standing wave, also known as a stationary wave.

2. What affects the shape of a stationary wave pattern?

The shape of a stationary wave pattern is affected by the frequency and amplitude of the vibration, as well as the properties of the medium (such as its tension and density). Changing any of these variables can alter the shape of the pattern.

3. How can I maintain a stable stationary wave pattern?

To maintain a stable stationary wave pattern, you will need to ensure that the frequency and amplitude of the vibration remain constant. Any changes to these variables can cause the pattern to shift or dissipate.

4. Can a stationary wave pattern be maintained indefinitely?

No, a stationary wave pattern cannot be maintained indefinitely. Eventually, the energy from the vibration will dissipate and the pattern will disappear. However, the pattern can be maintained for longer periods of time if the medium has low friction and the vibration is strong enough.

5. How can I change the position of a stationary wave pattern?

To change the position of a stationary wave pattern, you can either change the frequency or amplitude of the vibration, or change the properties of the medium (such as its tension or density). These changes will cause the pattern to shift and form in a different location.

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