The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

Here,
= Linear mass density of the string
Angular frequency of the wave on the string
A = Amplitude of the wave
v = Speed of the wave
At the same time each of this terms have its own definition, i.e,
Here T is the Period
For the linear mass density we have that

And the angular frequency can be written as

Replacing this terms and the first equation we have that



PART A ) Replacing our values here we have that


PART B) The new amplitude A' that is half ot the wavelength of the wave is


Replacing at the equation of power we have that

