Question

A transverse wave is travelling on a string with velocity $v$. The shape of string at $t=1s$ is given by $y=\frac{5}{x^2+6x+9}$ and at $t=2s$, it is given by $y=\frac{5}{x^2+12x+36}$, then fill the value of $(v+c)$, where $c$ denotes the direction of motion of wave ($+1$ for positive $x$-direction $\&-1$ for $-vex$-direction).

Answer 1

Answer: 2

$y(\text{ at }t=1)=\frac{5}{x^2+6x+9}$
[ by assuming wave is moving in +ive $x$ direction]
$y(x,t)=\frac{5}{\left((x-v(t-1){)}^2+6(x-v(t-1))+9\right)}$
$y($ at $t=2)=\frac{5}{(x-v{)}^2+6(x-v)+9}$
$=\frac{5}{x^2+x(6-2v)+v^2-6v+9}$
by comparing with given equation at $t=2$
$6-2v=12$
$\Rightarrow v=-3m/s$
$\Rightarrow$ -ive $\times$ -direction
$\Rightarrow$ speed $=3m/s$
$\Rightarrow$ Ans. $V+C=3+(-1)=2$