UNLPHY21 is EC150SPRING2017 Chapter 15 Homework Problem 15.70 Problem 15.70 A st

Question

UNLPHY21 is EC150SPRING2017 Chapter 15 Homework Problem 15.70 Problem 15.70 A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 191 m/s and a frequency of 240 Hz The amplitude of the standing wave at an antinode is 0.420 cm ngphysics.com Session maste Help I Close Resources previous l 8 of 8 I return to assignment Part A Calculate the amplitude at point on the s a distance of 17.0 cm from the left-hand end of the string. A- 0.42 10 m Submit My Answers Give Up incorrect; Try Again; 9 attempts remaining Not quite. Check through your calculations, you may have made a rounding error or used the wrong number of significant figures. Part B How much time does it take the string to go from its largest upward displacement to its largest downward displacement at this point? Submit My Answers Give Up

Explanation / Answer

part A:

frequency=f=240 Hz

amplitude=A=0.42 cm

wavelength=lambda=speed/frequency=0.79583 m

left end is a node i.e. amplitude of th wave is 0.

then the equation of oscillation as a function of distance x from the left end

=A*sin((2*pi/lambda)*x)

=0.42*sin(7.8951*x)cm

at x=17 cm=0.17 m,

amplitude=0.42*sin(7.8951*0.17) cm

=0.40907 cm

=0.40907*10^(-2) m

part B:

time taken=0.5*time period

=0.5*(1/frequency)

=0.0020833 seconds

part C:

maximum velocity A*2*pi/lambda

=0.033159 m/s

part D:

maximum acceleration=(2*pi/lambda)^2*A

=0.2618 m/s^2

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