Solving Moving Crests Problem: Acceleration Zero @ .75m

  • Thread starter imranq
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In summary, the question asks at what point on a 3 m long string with two periods will the crest have an acceleration of zero. The conversation discusses the various interpretations and equations used to solve the problem, but ultimately there is no clear solution due to the unclear wording of the question.
  • #1
imranq
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Homework Statement


A crest is created on a string by vibrating it. The crests have two periods on the string which is 3 m long. At what point will the crest have an acceleration of zero?

I had gotten this problem from a science competition and really had no idea how to do it. But I assumed that the vibrations would make a sine curve and the place where acceleration is zero would be where [tex] \frac {d^{2}}{dx^{2}} \sin {x} = 0 [/tex]

Thereby I calculated an answer of .75m, but could someone clarify this for me?

Homework Statement

Homework Equations


The Attempt at a Solution


.75m
 
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  • #2
I can't see that there is a sensible answer to this question. The 'crest' of a wave usually moves forward at a steady velocity. This implies no acceleration. Perhaps the question refers to a stationary wave. The crest then never moves!
How a crest has two periods does not make sense to me either. Does this mean there are two crest on the string (stationary wave) or one crest takes two time periods to traverse the string(progressive wave)
Alternatively you could consider the acceleration back to the equilibrium position ie from the crest down to the mid line. But a crest is the point of max acceleration by definition.
 
  • #3
I think when they mention 'crest' they mean to imagine a very small amount of string that is moving in a crest and that obviously accelerates and decelerates depending on how the crest is created. That was my logic, or else I couldn't solve it.
 
  • #4
Ok If we concentrate on one piece of the string then it is in simple harmonic motion.

The equation for this is a=-Amplitude*(2*pi*freq.)^2*sin[(2*pi*freq.)t]
a= acceleration
amplitude = max distance moved from mid point
freq= number of oscillations per second

But this does not seem to get a solution any closer. There are too many unknowns.

I'm still unclear as to whether they mean a standing wave like the vibration of a guitar string or a progressive wave that moves along the string.

Have you got the precise wording of the question correct?
 
  • #5
Its hazy right now, maybe the question had a different wording, but I can't check right now
 

Related to Solving Moving Crests Problem: Acceleration Zero @ .75m

1. What is the "Solving Moving Crests Problem"?

The "Solving Moving Crests Problem" refers to a scientific problem that involves finding the acceleration of a moving object when it reaches a crest at a specific distance of .75 meters.

2. Why is the acceleration zero at .75 meters?

The acceleration is zero at .75 meters because at this point, the moving object has reached the peak of its motion and is about to change direction. Therefore, it experiences a momentary pause in its acceleration before accelerating in the opposite direction.

3. What factors affect the acceleration of the moving object at .75 meters?

The acceleration of the object at .75 meters is affected by various factors such as the initial velocity of the object, the mass of the object, the force acting on the object, and the presence of any external forces like air resistance or friction.

4. How can the "Solving Moving Crests Problem" be applied in real life?

The "Solving Moving Crests Problem" has practical applications in various fields such as engineering, physics, and sports. For example, it can be used to calculate the acceleration of a roller coaster at its highest point or to determine the force required to launch a projectile to a certain height.

5. What are the steps involved in solving the "Solving Moving Crests Problem"?

The steps involved in solving the "Solving Moving Crests Problem" include identifying the initial velocity, determining the mass of the object, calculating the net force acting on the object, and using the equation of motion to find the acceleration at .75 meters. These steps can be repeated for different distances to get a complete understanding of the motion of the object.

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