Help Solving Harmonic Motion Problem

In summary: MENDED:In summary, the conversation is about finding the total force exerted on a 1/2 wavelength section of rope carrying a specific wave, taking into consideration small-angle approximation and neglecting the weight of the rope. The force is determined by finding the vector sum of the tensions at each end of the rope, which can be calculated by determining the slope of the rope at each end. The speed of the wave can be found from the wave function and used to calculate the tension.
  • #1
justagirl
29
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Can you offer some help on the following question? I've been trying different things for a while but can't get the right answer. Thanks!

At time t = 0, consider a 1/2 wavelength long section of the rope which is carrying the wave y = 0.06 cos(2.1 t - 2.4 x) between two points which have zero displacement (y = 0). Find the total force exerted by the rest of the rope on this section. Neglect any effects due to the weight of the rope. Use the small-angle approximation where q, sin(q), and tan(q) are all approximately equal to each other. A 5 meter length of this rope has a mass of 1.5 kg.

What I have so far:
I know the horizontal components of the Tensions cancel out, so the net force is due to the Y component of the one of the forces times 2. How do I find that?

Thanks a lot!
 
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  • #2
justagirl said:
At time t = 0, consider a 1/2 wavelength long section of the rope which is carrying the wave y = 0.06 cos(2.1 t - 2.4 x) between two points which have zero displacement (y = 0). Find the total force exerted by the rest of the rope on this section. Neglect any effects due to the weight of the rope. Use the small-angle approximation where q, sin(q), and tan(q) are all approximately equal to each other. A 5 meter length of this rope has a mass of 1.5 kg.

The force on the rope is the vector sum of the tensions at each end. Since horizontal tension T is constant,

[tex]T_1 = T/cos\alpha_1 \text{ and } T_2 = T/cos\alpha2[/tex]

You can determine the slope of the rope at each end (dy/dx) by partial differentiation of the wave function to find [itex]\alpha_1 \text{ and } \alpha2[/itex] where y = 0 (ie. [itex]2.1 t - 2.4 x = \pi/2, 3\pi/2[/itex])

You should be able to determine the speed of the wave from the wave function using the general solution:
[tex]y = Asin(2\pi/\lambda (x - \nu t))[/tex]

to the wave equation for the ideal string:

[tex]\frac{\partial ^2y}{\partial x^2} = \frac{\rho}{T}\frac{\partial ^2y}{\partial t^2}[/tex]

and then use
[itex]v = \sqrt{T/\rho} = \lambda \nu [/tex] to find the Tension T.

AM
 
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  • #3


Sure, I'd be happy to offer some help with this problem. First, let's review the given information: we have a 1/2 wavelength long section of rope carrying a wave with a displacement function of y = 0.06 cos(2.1 t - 2.4 x). The wave is traveling between two points with zero displacement, and we are asked to find the total force exerted by the rest of the rope on this section. We are also given the mass of a 5 meter length of this rope (1.5 kg).

To solve this problem, we will need to use the equation for the force exerted by a wave on a rope, which is F = μω^2A, where μ is the linear mass density (mass per unit length) of the rope, ω is the angular frequency of the wave, and A is the amplitude of the wave.

First, we need to find the linear mass density of the rope. We are given the mass of a 5 meter length of the rope (1.5 kg), so we can use this to find the mass per unit length: μ = 1.5 kg / 5 m = 0.3 kg/m.

Next, we need to find the angular frequency of the wave. We can use the equation ω = 2πf, where f is the frequency of the wave. Since we are given the equation for the wave (y = 0.06 cos(2.1 t - 2.4 x)), we can see that the frequency is 2.1 Hz. Therefore, ω = 2π(2.1) = 4.2π rad/s.

Finally, we need to find the amplitude of the wave. The amplitude is the maximum displacement of the wave, which in this case is 0.06 meters.

Now, we can plug these values into the equation for force: F = (0.3 kg/m)(4.2π rad/s)^2 (0.06 m) = 0.075π N.

Since the wave is traveling in both directions, the total force exerted by the rest of the rope on this section is twice this amount, or 0.15π N.

I hope this helps! Let me know if you have any further questions.
 

FAQ: Help Solving Harmonic Motion Problem

1. What is harmonic motion?

Harmonic motion is a type of periodic motion in which an object moves back and forth around an equilibrium point, with a restoring force that is directly proportional to the displacement from the equilibrium point.

2. How do you solve a harmonic motion problem?

To solve a harmonic motion problem, you need to first identify the relevant variables such as amplitude, frequency, and initial position. Then, you can use equations such as the harmonic motion equation or the energy conservation equation to find the unknown variables.

3. What is the difference between simple harmonic motion and damped harmonic motion?

Simple harmonic motion occurs when there is no external force acting on the object, while damped harmonic motion occurs when there is a resistive force, such as friction, acting on the object. This results in a decrease in amplitude over time in damped harmonic motion.

4. Can you use calculus to solve harmonic motion problems?

Yes, calculus can be used to solve harmonic motion problems, especially for more complex systems. The position, velocity, and acceleration of an object in harmonic motion can be described by trigonometric functions, which can be differentiated and integrated using calculus.

5. What are some real-life examples of harmonic motion?

Some common examples of harmonic motion include the swinging of a pendulum, the vibrations of a guitar string, and the motion of a mass attached to a spring. Other examples include the motion of a child on a swing, the movement of a bungee jumper, and the oscillations of a diving board.

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