Does the equation
`y=+sqrt(16-(2x-t)^(2)`,
`(2x-t)lt4 or (2x-t)gt-4`,
=0,otherwise
represent a wave? If yes, then find the amplitude and the phase velocity.

Explanation

Evidently, the factor, consistanig of `x` and `t`, appear as a combination of `[(x//y)-t]` namely `v=1//2` unit. Further, the condition `(2x-t)lt4`
and `(2x-t)gt-4`, can be equivalently expressed as `|2x-t|^(2)lt4^(2)` or, `16-(2x-t)^(2)gt0`
therefore, the function `y` is define for the restricted value of `x` and `t` . At the same time, the function is also bounded. hence, the given equation represents a wave,
The amplitude of the wave will be the maximum displacement `y` of the oscillating particle.
Therefore, for `y` to be maximum `(2x-t)^(2)` should be minimum, since the minimum value of this square can be zero, so force towards equilibrium position `(kx' sin theta) =dk(tantheta-(3sintheta)/4)` due to one spring and `net=2dk(tan theta-(3theta)/4)` for small `theta`,force `=2dk[theta-(30)/(4)]=k((d theta)/(2))`
`d theta=`displacement from mean position `impliesF-(kx)/(2) rArr T_(B)=2pisqrt((2m)/(k))`
`rArr T_(b)/T_(A)=2pisqrt(m/2k)//2pisqrt(2m/k)`
`rArrT_(B)/T(A)=2`
`(2x-t)^(2)_(min.)=0 :.y_(max)=4 unit`
Thus, the amplitude is `4` unit and the phase velocity ia `(1//2)` unit.