Question

The displacement equation for a particle in standing wave is represented as $y(x,t)=0.4\sin \left(0.5x\right)\cos \left(30t\right)$, where $x$ and $y$ are in $\text{cm}$.
Magnitude of velocity of particle at $x=2.4\text{cm}$ and $t=0.8\text{s}$ is -
$\left[\sin \right(1.2)=0.93,\text{sin}(24)=-0.9]$

• A. $0.4\text{cm/s}$
• B. $5.2\text{cm/s}$
• C. $10.1\text{cm/s}$
• D. $12\text{cm/s}$

Answer 1

The correct option is C $10.1\text{cm/s}$

Given:
$y(x,t)=0.4\sin \left(0.5x\right)\cos \left(30t\right)$
On differentiating above equation partially w.r.t time $t$, i.e. keeping $x$ constant.
$\Rightarrow \overrightarrow{v}(x,t)=\frac{\delta y}{\delta t}=-12\sin \left(0.5x\right)\sin \left(30t\right)$
On putting $x=2.4\text{cm}$ and $t=0.8\text{s}$
$\Rightarrow \overrightarrow{v}=-12\sin (0.5\times 2.4)\sin (30\times 0.8)$
$\Rightarrow \overrightarrow{v}=-12\times 0.93\times (-0.9)$
$\Rightarrow \overrightarrow{v}=10.1\text{cm/s}$
So, speed of the particle is $10.1\text{cm/s}$