Consider a wave sent down a string in the positive direction whose equation is given is
`y=y_(0)sin[omega(t-x/v)]`
The wave is propagated along because each string segment pulls upward and downward on the segment adjacent to it a slightly larger value of `x` and, as a result does work upon the string segment to which wave is travelling.
For example, the portion of string at point `A` is going upward, and will pull the portion at point `B` upward as well. In fact, at any point along the string, each segment of the string is pulling on the segment just adjacent and to its right, causing the wagve to propagate. It is by this process that the energy is sent along the string.
Now we try to calculate how much energy is propagated down the string per second
`T_(y)=tsintheta~~TtanthetaimpliesT_(y)=-T(dely)/(delx)`
(The negative sign appears because as shown in the figure II, the slope is negative) the force will act through a distance
`dy=v_(y)dt=(dely)/(delt) dt`
Therefore work done by force in time `dt` is
`dW=T_(y)dy=-T((dely)/(delx))((dely)/(delt))dtimpliesdW=(omega^(2)y_(0)^(2)T)/vcos^(2)[omega(t-x/V)]dt`.....(A)

How much average power is transmitted dawn a string having density `mu=5xx10^(-4)kg//m` and `T=5N` by a `200Hz` vibration of amplitude of `0.20cm`

Options

A. `0.08` Watt
B. `0.16` Watt
C. `0.32` Watt
D. `0.64` Watt

Answer

B

Explanation

Since the work done will be same at any place along the string in one cycle so choose `x=0`, for convenience. Then work done in a cycle `=(omega^(2)y_(0)^(2)T)/v int_(0)^(2pi//omega) cos^(2) omegat dt=(piomegay_(0)^(2)T)/v`
Average power `=` work done per unit time `=("work done in a cycle")/("Time period")=((piomega_(0)^(2)T).v)/((2pi)/omega)=(omega^(2)y_(0)^(2)T)/(2V)`
Average power `=2pi^(2)f^(2)y_(0)^(2)T sqrt((mu)/T)=2pi^(2)f^(2)y_(0)^(2)sqrt(muT)`
`=2xx10xx200xx200xx0.2xx0.2xx10^(-4)sqrt(5xx10^(-4)xx5)=160xx10^(-3)` Watt `=0.16` Watt.