Question

A heavy rope 6.00 mm long and weighing 23.4 N is attached at one end to a ceiling and hangs vertically. A 0.590-kg mass is suspended from the lower end of the rope.

Part A)

What is the speed of transverse waves on the rope at the bottom of the rope?

Part B)

What is the speed of transverse waves on the rope at the middle of the rope?

Part C)

What is the speed of transverse waves on the rope at the top of the rope?

Part D)

Is the tension in the middle of the rope the average of the tensions at the top and bottom of the rope? Is the wave speed at the middle of the rope the average of the wave speeds at the top and bottom? Select the correct answer and explanation. The tension at the middle is the average of the top and bottom tensions, but the average speed is not equal to the speed at the middle because the speed depends on the square root of the tension. The speed at the middle is the average of the top and bottom speeds, but the average tension is not equal to the tension at the middle because the tension depends on the square root of the speed. The speed at the middle is not equal to the average speed, and the average tension is not equal to the tension at the middle because the tension is proportional to the speed. o The tension at the middle is the average of the top and bottom tensions, and the speed at the middle is the average of the top and bottom speeds because the speed is proportional to the tension.

Answer 1

Draw the diagram of the given situations of the quests -oh calculati the mass fu mit feugth of the ropes. Have = 23.4. = 397.959 kgim Let, y= length of the lower part of the rope T= tension force acting upward egy = clownward force mg = mass suspended. equation will be written as, o FT + = Т - А) speed of the trausnesse wave is given by v=5Tu = Hygting = 1oy-t meg a substituts the given ratues from question. V= /19-8)(y) +(0.590) 21:02 = -8y +0.014529 397.959 A) at botton of rofe iy=0 Va 19.01x0 +0.014529 = 0.12053 mis B) at widelly, Y = 2 2 = 6x10 3 = 3x10 m vs 19.01)(3418) +0.14522 = 0.417.99 mio. शाशEETEENETI 2

E c) at top 7=2= 6x10 m = √ (6x103) (9.8) +0:014529 20.27079 m/s from egno average tension will be T (LA = Mig ting -(tension is midelle) to tang = Mig ting Caverage seunion) 1. Till = 2 Mig + mg (tension in top so, the tension teation in midell is equal to the average tension of top and bottom. But, speed at midelle is not equal to average sfeed of bottom and top. so, Ist statement is right choice. SPISETEE537252525252525252