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timama [110]
3 years ago
13

Adjacent antinodes of a standing wave on a string are 15.0 cmapart. A particle at an antinode oscillates in simpleharmonic motio

n with amplitude 0.0850 cm and period 0.0750 s. Thestring lies along the +x-axis and is fixed at x=0.
A) How far apart are the adjacent nodes?
B) What is the wavelength of the two traveling waves that formthis pattern?
C) What is the amplitude of the two taveling waves that formthis pattern?
D) What is the speed of the two traveling waves that form thispattern?
E) Find the maximum and minimum transverse speeds of a pointat an antinode.
F) What is the shortest distance along the stringbetween a node and antinode?
THANKS!
Physics
1 answer:
viva [34]3 years ago
5 0

Answer:

a)   D = 15 cm , b)  λ = 30.0 cm , c) 0.0850 cm, d) v = 400 cm / s , e)    v = 80.43 cm / s , v = -80.43 cm / s, f)   D₂= 7.5 cm

Explanation:

Standing waves form when two waves of the same frequency travel in opposite directions,

A) in the waves the distance of the nodes and antinode are the same since the wavelength is constant

          D = 15 cm

B) the wavelength is the distance for which the wave repeats itself, in the case of a standing wave, the distance between two nodes is lamita of the wavelength.

        D = λ / 2

         λ=  15 2

         λ = 30.0 cm

C) the amplitude of each wave is 0.0850 cm, the amplitude of the standing wave is double A = 0.17 cm

D) Let's use the speed ratio

        v =  λ f

        f = 1 / T

        v =  λ / T

       v = 30.0 /0.0750

       v = 400 cm / s

E) the transverse speeds are the speed of the oscillatory movement

       y = A cos (wt)

 

      w = 2π f = 2π / T

      w = 2π / 0.0750

      w = 83.78 rad / s

     

      y = 0.850 cos (83.78 t)

Speed ​​is

      v = dy / dt

      v = -A w cost wt

      v = - 0.850 83.78 cos (83.78 t)

      v = -80.43 cos (83.78 t)

The maximum speed when the cosine values ±1

       v = 80.43 cm / s

       v = -80.43 cm / s

F) if we draw a drawing, the distance between two nodes is half the wavelength, at the distance between an antinode synod is half this, it occupies a quarter of the wavelength

      D₂ = ¼  λ

      D₂ = 30.0/4

      D₂= 7.5 cm

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