Question #2f8d7

1 Answer
Aug 20, 2015

The wavelength should decrease by #9 1/11%#

Explanation:

Assuming the velocity of wave propagation remains constant.

Then
#color(white)("XXXX")v=w*f# is a constant
where #v# is the velocity, #w# is the wavelength, and #f# is the frequency

If #w# is increased by #10%# giving new wavelength, #hatw=1.1*w#
then we will need a new frequency #hatf#

#color(white)("XXXX")w*f = hatw*hatf = 1.1w*hatf#

#rArrcolor(white)("XXXX")hatf=1/1.1f =10/11f #

#rArrcolor(white)("XXXX")f- 10/11f = 1/11f# decrease

#1/11 = 9.090909...%#