Question

A wave travelling along a string is described by the equation $y=A\,sin\,\left(\omega t-kx\right)$ The maximum particle velocity is

• A. $A\omega$
• B. $\omega/k$
• C. $d\omega/dk$
• D. $x/l$

Answer 1

Answer: (A)

$A\omega$

Answer 2

Given that, the displacement of a particle is
$y=A \sin (\omega t-k x)$...(i)
The particle velocity
$v_{p}=\frac{d y}{d t}$...(ii)
Now, on differentiating E (i) w.r.t, $t$
$\frac{d y}{d t}=A \cos (\omega t-k x) \cdot \omega$
$\Rightarrow \frac{d y}{d t}=A \omega \cos (\omega t-k x)$
From E (ii)
$\Rightarrow v_{p}=A \omega \cos (\omega t-k x)$
For maximum particle velocity,
$\cos (\omega t-k x)=1$
So, $v_{p}=A \omega \times 1$
$\Rightarrow v_{p}=A \omega$