Calculating Tension in a Stretched Rope: A Harmonic Wave Example

[Rockdog]

The transverse displacement of a harmonic wave on a stretched rope is y = 0.03 cos(2.7 t - 2.9 x), where x and y are in meters and t is in seconds. A 5 meter length of this rope has a mass of 1.5 kg.
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a) What is tension in rope?
b) At time t = 0, consider a 1/2 wavelength long section of the rope which is carrying the wave y = 0.03 cos(2.7 t - 2.9 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.

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a) Not too bad. Tension= V^2*mu
where V is velocity
and mu is linear density. V=omega/K, thus, v=.9310 m/s
mu=mass/length...so 1.5kg/5m =.2600N

Tension equals .2600N, which is correct answer.

b) Ok, part b is giving me a headache. I don't know how to start this problem really. I've included a picture to help.

 

[member 553083]

The total force exerted by the rope on the 1/2 wavelength section can be calculated using the equation F=T*sin(theta), where T is the tension in the rope and theta is the angle between the two ends of the section. Using the small-angle approximation, we can assume that sin(theta) = theta, where theta is the angle in radians. Thus, the total force exerted by the rope on the 1/2 wavelength section is F = T*theta. Substituting in the values for T and theta, we get F = 0.2600N * (2.9m / 5m) = 0.1548N. This is the correct answer.

 

[M98Ranger]

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b) To find the total force exerted by the rest of the rope, we can use the equation F=ma, where F is the force, m is the mass, and a is the acceleration. In this case, the acceleration is given by the second derivative of the displacement function, which is -0.2379 cos(2.7t - 2.9x). Since we are only considering a half-wavelength section of the rope, the mass can be calculated by multiplying the linear density (mu) by the length of the half-wavelength (L/2). Thus, m= (1.5kg/5m)*(2.7m/2) = 0.405kg.

Plugging in these values, we get F= (0.405kg)(-0.2379cos(2.7t-2.9x)) = -0.0965cos(2.7t-2.9x).

Using the small-angle approximation, we can simplify cos(2.7t-2.9x) to approximately 1, since the angle is small. Therefore, the total force exerted by the rest of the rope on this section is approximately -0.0965N.

We can also verify this result by using the equation F= Tsin(q), where T is the tension and q is the angle between the tension and the horizontal direction. Since the section of rope is at zero displacement, the tension in the rope must be equal to the total force exerted by the rest of the rope. Using the small-angle approximation again, we get T= F/sin(q) = -0.0965N/sin(0) = -0.0965N.

Therefore, the total force exerted by the rest of the rope on this section is approximately -0.0965N, which is the same as our previous result.

In conclusion, the total force exerted by the rest of the rope on this half-wavelength section is approximately -0.0965N, which is the same as the tension in the rope. This shows that the tension in the rope is responsible for the acceleration of the wave on the rope.