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.(adsbygoogle = window.adsbygoogle || []).push({});

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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.

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# Homework Help: Transverse Waves Help

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