os//edugen wileyplus.com/edugen/iti/main.uni Halliday, Fundamentals of Physics,

Question

os//edugen wileyplus.com/edugen/iti/main.uni Halliday, Fundamentals of Physics, 10e Hele I System Announcements (1 Unread) PR?NTER VERS?ON (?? Chapter 16, Problem 050 For a certain transverse standing wave on a long string, an antinode is at x - o and an adjacent node is at x - o transverse velocity of the string particle at x - 0.20 m at (c)t 0.80 s and (d) t 1.4 5 he y axis is set by ys 4,2 cm. When t-0.80 s, what is the displacement of the string particle at .40 . The displacement y of the stri partide atx-8 shom? m a) X O.20 m and (b) x-0.70 m ? whats the (a) Number (b) Number (c) Number (d) Number Units Units Units Units TO SAMPLE PROPLEN DEO HIN?-LECTURE V?OSO HINT Question Attempts: 0 of 6 used

Explanation / Answer

equation for standing wave is given by:

y=2*A*cos(2*pi*x/lambda)*sin(w*t+theta)

where A=amplitude

lambda=wavelength

w=angular frequency

theta to be determined

at x=0,

from the equation:

y=2*A*sin(w*t+theta)

from the given graph:

amplitude=4.2 cm

time period=2 seconds

==>w=2*pi/time period=3.1416 rad/s

hence y=-4.2*sin(3.1416*t)=4.2*sin(3.1416*t+pi)

comparing with the given equation,

2*A=4.2==>A=2.1 cm

theta=pi

w=3.1416 rad/s

also given that, x=0 is an antinode and x=0.4 is a node.

node means amplitude of y is always 0 irrespective of time

hence cos(2*pi*0.4/lambda)=0

==>2*pi*0.4/lambda=pi/2

==>lambda=1.6 m

so complete standing wave equation:

y=4.2*cos(2*pi*x/1.6)*sin(3.1416*t+pi)

part a:

t=0.8 s, x=0.2 m

y=4.2*cos(2*pi*0.2/1.6)*sin(3.1416*0.8+pi)

=-1.7456 cm

part b:

t=0.8 s, x=0.7 m

y=4.2*cos(2*pi*0.7/1.6)*sin(3.1416*0.8+pi)

=2.2808 cm

part c:

velocity=dy/dt

=4.2*cos(2*pi*x/1.6)*3.1416*cos(2.618*t+pi)

t=0.8 s, x=0.2 m

velocity=4.2*cos(2*pi*0.2/1.6)*3.1416*cos(3.1416*0.8+pi)

=7.5482 cm/s

part d:

t=1.4 s, x=0.2 m

velocity=4.2*cos(2*pi*0.2/1.6)*3.1416*cos(3.1416*1.4+pi)

=2.8831 cm/s

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